The rate at which the surface area of the sphere is increasing when the radius of the sphere is 12 cm is 1/226.5 cm/s.
What is the rate of surface area change?The volume of a sphere is increasing at the rate of 8 cm³/s.
Radius of the sphere is 12 cm.
So, we need to find the rate at which its surface area is increasing.
Let, V be the volume of the sphere and r be the radius of the sphere. The volume of a sphere of radius r is given by:
V = (4/3)πr³
Differentiating with respect to time t, we get:
dV/dt = 4πr²(dr/dt) ...(1)
Also, the surface area of the sphere is given by:
A = 4πr²
Differentiating with respect to time t, we get:
dA/dt = 8πr(dr/dt) ...(2)
From equations (1) and (2), we can write:
dr/dt = dV/dt ÷ 4πr²
dr/dt = 8 / (4π × 12²)
dr/dt = 8 / 1808
dr/dt = 1 / 226.5 cm/s
Therefore, the rate at which the surface area of the sphere is increasing when the radius of the sphere is 12 cm is 1/226.5 cm/s.
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A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). The ratio r3T2 for the moon is 1.01×1018km3y2. Calculate the radius of the orbit of such a satellite. All work must be shown for full credit. The choices are: 2.75x10E3 km; 1.96x10E4km; 1.40x10E5km; 1.00x10E6km.
The radius of the orbit of such a satellite will be about 1.40 × 10⁵ kilometers.
What is the radius of orbit?To calculate the radius of the orbit of a geosynchronous Earth satellite, we must use the equation:
r³T² = 1.01 × 10¹⁸ km³y²
where, r is the radius of the orbit and T is the orbital period of the satellite, which is 1 day. We can rearrange the equation to calculate r, giving us:
r = (1.01 × 10¹⁸km³y²)1/3/(1 day)2/3
To calculate the radius of the orbit, we need to convert the units of 1 day to seconds: 1 day = 86400 seconds. We can substitute this into the equation:
r = (1.01 × 10¹⁸km³y²)1/3/(86400 seconds)2/3
Finally, we can calculate the radius of the orbit: r = 1.40 × 10⁵ km
Therefore, the radius of the orbit will be about 1.40 × 10⁵ km.
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When the price of radios decreases 5%, quantity demanded increases 5%. The price elasticity of demand for radios is ________ and total revenue from radio sales will ________.
Price elasticity of demand for radios is 1 and total revenue from radio sales will remain constant.
Price elasticity of demand is calculated as the percentage change in quantity demanded divided by the percentage change in price. Using this formula, we can calculate the price elasticity of demand for radios as follows:
Price elasticity of demand = (percentage change in quantity demanded) / (percentage change in price)
Given that when the price of radios decreases by 5%, quantity demanded increases by 5%.So, the percentage change in quantity demanded = 5% and the percentage change in price = -5%. (Because price has decreased by 5%.)Price elasticity of demand = (5% / -5%) = -1.The negative sign indicates that the demand is elastic. However, the question asks for a positive value, so we take the absolute value of -1.Price elasticity of demand = 1.
Therefore, the price elasticity of demand for radios is 1.When the price elasticity of demand is equal to 1, it means that the demand is unit elastic. This implies that the percentage change in quantity demanded is equal to the percentage change in price. If the price of radios decreases by 5% and the quantity demanded increases by 5%, it means that the total revenue from radio sales will remain constant. In other words, the increase in quantity demanded is exactly offset by the decrease in price, resulting in the same total revenue.
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how do the summer and winter monsoon affect climate in the region?
The summer monsoon brings heavy rainfall and cooler temperatures, while the winter monsoon brings dry, cool air to the region.
The summer monsoon is characterized by winds blowing from the southwest over the Indian Ocean, bringing moisture to the Indian subcontinent and Southeast Asia. This results in heavy rainfall, cooler temperatures, and increased humidity during the summer months. The winter monsoon, on the other hand, is characterized by winds blowing from the northeast, bringing dry, cool air to the region, leading to lower temperatures and little to no rainfall. The seasonal changes brought by the monsoon winds play a crucial role in shaping the climate of the region, affecting everything from agriculture to water resources to human settlements.
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Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 22 minutes. Determine the drag force on the runner during the race. Suppose that the cross section area of the runner is 0.72 m2 and the density of air is 1.2 kg/m3.I know how to get the drag force, but have no idea how to get the drag coefficient, in order to plug into the equation! I found the velocity in m/s, then went to find the force using F=1/2(density of air)(velocity^2)(drag coefficient)(cross section area) but don't know what to use for the drag coefficient.
Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 22 minutes. The drag force on the runner during the race is 13.4 N.
Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Drag force is a form of air resistance that acts on objects moving through air. When a runner is running on a treadmill, there is no drag force to work against.
In order to calculate the drag force on the runner during the race, we need to determine the drag coefficient. The drag coefficient is a dimensionless number that represents the ratio of drag force to dynamic pressure. It is affected by the shape and size of the object as well as the fluid (air) it is moving through. Generally, a higher drag coefficient means that more force is required to move the object.
To calculate the drag coefficient, we can use the following formula: Cd = Fd / (1/2 * ρ * v2 * A), where Fd is the drag force, ρ is the density of the air, v is the velocity of the object, and A is the cross-sectional area of the object.
For our example, we are given a runner that is 60 kg and completed a 5 km race in 22 minutes. The velocity of the runner can be calculated by v = d/t, where d is the distance traveled and t is the time taken. This gives us a velocity of 8.3 m/s. The density of the air is given to be 1.2 kg/m3 and the cross-sectional area is 0.72 m2.
Plugging these values into the formula gives us a drag coefficient of 0.385. This means that for every 1 unit of dynamic pressure, the drag force is 0.385. We can now calculate the drag force on the runner by multiplying the drag coefficient by 1/2 * ρ * v2 * A. In this case, the drag force is 13.4 N.
In conclusion, the drag force on the runner during the race is 13.4 N. This was calculated by determining the drag coefficient using the formula Cd = Fd / (1/2 * ρ * v2 * A) and then multiplying it by 1/2 * ρ * v2 * A.
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If the change in internal energy = 1714J, specific
heat capacity = 49J/°C/kg, and mass = 38kg,
what is the temperature change experienced?
Give your answer to 2 decimal places.
Answer:
0.92°C
Explanation:
C = change in Q/m × change in T
so
change in T = change in Q/C ×m
C= 49
m= 38
change in Q= 1714
then
= 1714/49 × 38
= 1714/1862
= 0.92°C
rounded off to 2 d.p
if the variable capacitor in an fm receiver ranges from 10.9 pf to 16.4 pf , what inductor should be used to make an lc circuit whose resonant frequency spans the fm band?
To create an LC circuit spanning the FM band with a variable capacitor of 10.9-16.4 pF, use the formula L = 1/(4π²f²C).
The inductor needed to make an LC circuit whose resonant frequency spans the FM band depends on the variable capacitor in the FM receiver. In your case, the variable capacitor ranges from 10.9 pF to 16.4 pF. To determine the inductor needed for the LC circuit, you can use the following formula:
L = (1/ (4π² * f² * C))
Where:
"L" is the inductor. "f" is the frequency of the LC circuit. "C" is the capacitor.For example, if you set the variable capacitor to 10.9 pF, the inductor needed to make an LC circuit whose resonant frequency spans the FM band would be:
L = (1/ (4π² * f² * 10.9 pF))
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Complete the following sentence.
A diameter is also a...
Answer:
A diameter is also a double of radius
a satellite is shot into a low orbit around a newly discovered planet. if the satellite is traveling at 8400 m/s just above the surface, and the acceleration due to gravity on this planet is 14.4 m/s2 , what must be the planet's radius?
The planet's radius is approximately 2.13 × 10^6 meters.
Planet radius calculation.
To find the planet's radius, we can use the following formula:
v² = GM/r
where v is the satellite's velocity, G is the gravitational constant, M is the planet's mass, and r is the planet's radius.
Since the satellite is just above the surface of the planet, we can assume that r is equal to the sum of the planet's radius and the satellite's altitude above the surface. Let h be the altitude of the satellite above the planet's surface, then we have:
r = planet's radius + h
Substituting this expression for r into the equation above and solving for the planet's radius, we get:
r = GM/v² - h
where G = 6.6743 × 10^-11 Nm²/kg² is the gravitational constant.
Substituting the given values, we get:
r = (6.6743 × 10^-11 Nm²/kg²) * M / (8400 m/s)² - h
We can also use the formula for the acceleration due to gravity at the surface of a planet:
g = GM/r²
where g is the acceleration due to gravity at the planet's surface.
Solving for M in this equation, we get:
M = g * r² / G
Substituting the expression for r from above and solving for r, we get:
r = √(GM/g)
Substituting the given values, we get:
r = √((6.6743 × 10^-11 Nm²/kg²) * M / (14.4 m/s²))
Equating this expression for r with the previous one, we get:
(6.6743 × 10^-11 Nm²/kg²) * M / (8400 m/s)² - h = √((6.6743 × 10^-11 Nm²/kg²) * M / (14.4 m/s²))
Squaring both sides and rearranging, we get:
M = (8400 m/s)² * (14.4 m/s²) * h / (2 * G)
Substituting this expression for M into the equation for r, we get:
r = √((8400 m/s)² * h / (2 * g))
Substituting the given values, we get:
r = √((8400 m/s)² * h / (2 * 14.4 m/s²))
r = 2.13 × 10^6 meters
Therefore, the planet's radius is approximately 2.13 × 10^6 meters using v² = GM/r.
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a boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 7.2 m/s. if the coefficient of friction between the sled's runners and snow is 0.055 and the boy and sled together weigh 540 n, how far does the sled travel on the level surface before coming to rest?
The boy coasts down the hill on the sled and reaches a level surface with a speed of 7.2 m/s. The coefficient of friction between the sled's runners and the snow is 0.055, and the boy and sled together weigh 540 N. To determine how far the sled will travel on the level surface before coming to rest, we need to calculate the distance using the formula Distance = (Speed x Time) - (1/2 x Acceleration x Time2). We can determine the time it takes for the sled to come to rest using the equation Speed = (Friction x Normal Force) / Mass. So, Speed = (0.055 x 540N) / 540N = 0.055 m/s. Time = 7.2/0.055 = 130.9 seconds. We can then calculate the distance as Distance = (7.2 x 130.9) - (1/2 x 0.055 x 130.92) = 927.9 m. Therefore, the sled will travel 927.9 m before coming to rest.
The boy and sled weigh 540 N. A boy coasts down a hill on a sled, reaching a level surface at the bottom with a speed of 7.2 m/s. The coefficient of friction between the sled's runners and snow is 0.055. How far does the sled travel on the level surface before coming to rest?
The distance traveled by the sled on the level surface before coming to rest is 72.22 meters.
What is friction?
Friction is a force that opposes motion. It is the friction between the sled's runners and the snow that causes the sled to stop. The formula for frictional force is:f = μN where f is the frictional force, μ is the coefficient of friction, and N is the normal force, which is equal to the weight of the sled and boy since they are on a level surface. The normal force is given by: N = m*g where m is the mass of the sled and boy and g is the acceleration due to gravity, which is equal to 9.81 m/s^2.
How far does the sled travel on the level surface before coming to rest?
The sled will travel a certain distance, d, before it stops. The distance, d, is given by:d = (v^2 - u^2) / 2fwhere v is the final velocity, u is the initial velocity (which is 7.2 m/s), and f is the frictional force. The frictional force is f = μN = μmgSubstituting the given values:d = (7.2∧2 - 0∧2) / (2*0.055*540*9.81)d = 72.22 meters. Therefore, the sled will travel 72.22 meters on the level surface before coming to rest.
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a 6.96 nc charge is located 1.90 m from a 3.86 nc point charge. find the magnitude of the electrostatic force, in nano newtons, nn, that one charge exerts on the other.
The magnitude of the electrostatic force, in nano newtons, nn, that one charge exerts on the other is 57.54 nN.
The question needs to find out the magnitude of the electrostatic force, in nano newtons (nn), that one charge exerts on the other. Let us understand the given data before starting the solution.
Given data:
Charge 1 (q1) = 6.96 nCCharge 2 (q2) = 3.86 nCDistance between charges (r) = 1.90 mFormula used:
We use Coulomb's law to find the electrostatic force between the two charges.
Coulomb's Law
F = (k*q1*q2)/r²
Where,
F is the force between the charges,q1 and q2 are the two charges separated by a distance r,k is the Coulomb constant which is equal to 9 x 10⁹ Nm²/C²Let us substitute the given values in the above formula.
F = (9 * 10⁹) * (6.96 * 10⁻⁹) * (3.86 * 10⁻⁹) / (1.90)²F = 57.54 nN (nano newtons)Therefore, the magnitude of the electrostatic force, in nano newtons, nn, that one charge exerts on the other is 57.54 nN.
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the velocity v of an earth satellite varies directly as the square root of its mass m, and inversely as the square root of its distance r from the center of earth. if the mass is halved and the distance is doubled, how is the speed affected
If the mass of the satellite is halved and the distance is doubled, the velocity of the satellite will be reduced to approximately 70.7% of its original value.
What is a satellite in this context?
In this context, a satellite refers to an artificial object that is launched into orbit around the Earth to perform various functions, such as communication, navigation, and scientific research.
Let's start by writing the equation that relates the velocity of the satellite with its mass and distance from the center of the earth:
v = k√(m/r)
where k is a constant of proportionality.
Now, if the mass is halved and the distance is doubled, we have:
v' = k√(m/2r)
where v' is the new velocity. We can use this equation to find how the velocity is affected by the changes:
v' = k√(m/2r) = k√(m/r) / √2
The square root of 2 is approximately 1.414, so we can simplify the expression to:
v' = v / 1.414
Therefore, if the mass of the satellite is halved and the distance is doubled, the velocity of the satellite will be reduced to approximately 70.7% of its original value.
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Pensacola and Admiralty Head have very different tidal characteristics. Describe how a difference in location, shape of shoreline, and lunar declination likely contributes to the difference in tidal ranges and tidal patterns for these two locations.
The difference in location, shape of shoreline, and lunar declination likely contributes to the difference in the tidal ranges and tidal patterns for the two locations include landmasses and wave interaction.
What are tidal characteristics?The difference in tidal characteristics between Pensacola and Admiralty Head is likely due to the difference in location, shape of shoreline, and lunar declination. Location affects tidal ranges and patterns due to how different landmasses will interact with the waves.
The shape of the shoreline affects how the tides reflect and move in different directions. Lastly, lunar declination is a factor because the angle at which the moon is orbiting the earth affects the tides. This is because the gravitational pull of the moon varies with its distance and declination.
The differences in tidal characteristics between Pensacola and Admiralty Head can be attributed to the difference in location, shape of shoreline, and lunar declination, all of which have a direct impact on the tidal ranges and patterns of these two locations.
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if we are going to put a 36,000 btu/hr water heater and 120,000 furnace btu/hr ( both cat i appliances) in mechanical room that is 10' x 10' x 10' in size, what is the volume of space in the mechanical room?
The volume of the mechanical room is 1000 cubic feet
To calculate the volume of space in a mechanical room, given that a 36,000 btu/hr water heater and 120,000 furnace btu/hr (both cat i appliances) will be installed in a 10' x 10' x 10' room size, use the following formula:
Volume = Room Length x Room Width x Room Height
The volume of space in the mechanical room is given as follows:
Volume = 10' x 10' x 10'
Volume = 1000 cubic feet (cu ft)
Therefore, the volume of space in the mechanical room is 1000 cubic feet (cu ft).
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What is the difference between point to point encryption and end-to-end encryption?
Point-to-point encryption and end-to-end encryption are two distinct cryptographic approaches. Both these methods offer data security but in different ways.
The difference between point to point encryption and end-to-end encryption is as follows:
Point-to-point encryption
Point-to-point encryption (P2PE) protects payment card data from the time it is swiped to the point it is encrypted. It encrypts card data before it enters a merchant's system, keeping it secured until it is sent to the payment processor. The data is then decrypted and transmitted through the processing network to the card issuer for approval. P2PE prevents any attempts to intercept the card data while it's in motion from the terminal to the payment processor.
End-to-end encryption
End-to-end encryption (E2EE) involves encrypting data from the point of origin to its final destination. End-to-end encryption secures the entire data transmission process from client to server. It encrypts the data at the source, such that the data is protected throughout its journey. Therefore, end-to-end encryption is mainly used in messaging and communication apps like WA, etc.
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the diagram below shows four cannons firing shells with different masses at different angles of elevation. the horizontal component of the shell's velocity is the same in all four cases. in which case will the shell have the greatest range if air resistance is neglected? (a) cannon a (b) cannon b only (c) cannon c only (d) cannon d
The diagram below shows four cannons firing shells with different masses at different angles of elevation. the horizontal component of the shell's velocity is the same in all four cases. The case will the shell have the greatest range if air resistance is neglected is (a) cannon a
The cannon which would have the maximum range if air resistance is neglected is given by the expression R = (V²/g)sin(2θ). The horizontal component of velocity is the same for all four shells, Vx = Vcosθ. Where R is the range, V is the velocity, g is the gravitational acceleration, θ is the angle of projection, and Vx is the horizontal component of the velocity. The diagram below shows four cannons firing shells with different masses at different angles of elevation.
For the maximum range, we need to take the angle of projection to be 45°. The mass of the shell is not a consideration since it doesn't affect the time of flight or the range of the shell.Therefore, the maximum range is given by the highest value of V²sin(2θ)/g. As sin(90) = 1, sin(0) = 0, sin(30) = 1/2, sin(45) = √2/2, sin(60) = √3/2, sin(70) = 0.94, the maximum value of sin(2θ) is obtained when θ = 45°.For all four cannons, the horizontal component of velocity, Vx = Vcosθ, is the same. Therefore, the maximum range is obtained for Cannon A when air resistance is neglected. Therefore, the correct answer is (a) Cannon A.
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An object is propelled along a straight-line path by a force. If the net force were doubled, the object's acceleration would be? a) half as much b) twice as much c) the same d) none of these. e) four times as much.
An object is propelled along a straight-line path by a force. If the net force were doubled, the object's acceleration would be b. twice as much.
Force is a vector quantity that measures the interaction between two objects, it is described by its magnitude and direction. If there is no opposing force, the force will cause the object to accelerate. Acceleration is the rate at which the velocity of an object changes. The acceleration of an object is directly proportional to the force applied to it. So, if the net force acting on an object is doubled, the acceleration of the object will also double.
An object's acceleration is directly proportional to the net force acting on it, if the net force acting on an object doubles, the acceleration of the object will double as well. Force is a vector quantity that describes the interaction between two objects. The force is proportional to the product of the mass of an object and its acceleration. As a result, if the mass of an object is constant, the acceleration of the object will be directly proportional to the force applied to it. The relationship between force and acceleration is expressed in Newton's second law, which states that force equals mass times acceleration.
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imagine swinging a ball in a circle at the end of a string. if the string that holds the ball breaks, what causes the ball to move in a straight line path?
When a ball is swung in a circle at the end of a string, it is constantly changing direction due to the force acting on it. This force is called the centripetal force, which is provided by the tension in the string.
When the string holding the ball breaks, there is no longer any force acting on the ball to keep it moving in a circular path. As a result, the ball moves in a straight line path in accordance with Newton's first law of motion, which states that an object at rest will remain at rest or an object in motion will continue to move in a straight line path at a constant speed unless acted upon by an external force.
In this case, the external force was the tension in the string, which was providing the centripetal force to keep the ball moving in a circular path. Once the string broke, the ball no longer experienced any centripetal force, and thus continued to move in a straight line path.
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you are designing a spacecraft to a giant planet. which planet is your spacecraft going to study, and what is it going to learn about the planet?
A spacecraft is a vehicle that can travel into space. The spacecraft can be used to study other planets, asteroids, and comets in our solar system. Spacecraft has the ability to collect data, take photographs, and make measurements about the planets and other space objects.
What can you learn about a planet?With a spacecraft, scientists can learn a lot about planets. Some of the things that can be learned include the following:
The chemical composition of the planet's surface and atmosphere.The geology of the planet, such as mountains, valleys, and other features.How the planet rotates, and how long it takes to complete one rotation.The planet's weather patterns and climate, such as temperature and wind speeds.The planet's magnetic field, and how it interacts with the solar wind.The planet's moons and rings, and how they interact with the planet.In conclusion, with a spacecraft, scientists can learn a lot about planets. Information about a planet can vary depending on the planet.
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What planet rotates once a day?
Earth is the only planet with a daily rotation. The only planet in our solar system known to offer the ideal circumstances for supporting life is Earth, which is located third from the Sun.
The only planet in our solar system known to offer the ideal circumstances for supporting life is Earth, which is located third from the Sun. The rotation of our planet, which creates day and night, is one of its most striking characteristics. Every 24 hours, the Earth spins on its axis, giving rise to the cycle of day and night. The Coriolis effect, which affects the direction of winds, ocean currents, and other significant motions in the atmosphere and seas, is also a result of this rotation. The molten core of the globe spins as Earth rotates, creating a magnetic field that shields humans from dangerous solar radiation.
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(Figure 1) shows a collision between three balls of clay. The three hit simultaneously and stick together. Assume that m = 60 g and v = 2.9 m/s. ⬤↘ m 40 m/s, and 45°
←⬤ v 30 g
↑
⬤ 20 g and 2.0 m/s
Part A What is the speed of the resulting blob of clay? Express your answer with the appropriate units. V = ? Part B What is the movement direction of the resulting blob of clay? Express your answer in degrees below the horizontal. θ = ?
The speed of the resulting blob of clay is 20.99 m/s and the direction is 45.82⁰ below the horizontal.
Given :
Masses of balls of clay:
m₁ = 60g,
m₂ =20g,
m₃ = 30g.
Speed of balls of clay :
v₁ = 40m/s,
v₂= 2m/s,
v₃ = 2.9m/s
we can write the speed in vector form as :
υ₁ = 40( x + y)/ √2 m/s,
υ₂ = 2 y m/s,
υ₃ = 2.9 (-y) m/s, where x and y are unit vectors in perpendicular directions.
During a collision, the momentum remains conserved. Hence using the conservation of total momentum we can calculate the final speed of the resulting bob clay.
Using conservation of momentum,
initial momentum = final momentum
m₁υ₁ + m₂υ₂ + m₃υ₃ = (m₁+m₂+m₂)υ,
where υ = final velocity of clay blob.
Putting all the values in the above equation,
60 × 40( x + y)/ √2 + 20×2 y+30 ×2.9 (-y) = (60+20+30) υ
on solving the above equation, we get
υ = 14.63 x + 15.06 y
The magnitude of the final speed will be equal to √(14.63²+ 15.06²)
Final speed= 20.99 m/s.
and
Angle = tan⁻(15.06/14.63)
Angle = 45.82⁰ below the horizontal.
Therefore, the speed of the resulting blob of clay is 20.99 m/s and the direction is 45.82⁰ below the horizontal.
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Part A A canoe is designed to have very little drag when it moves along its length. Riley, mass 62 kg, sits in a 21 kg canoe in the middle of a lake. She dives into the water off the front of the canoe, along the axis of the canoe. She dives forward at 1.7 m/s relative to the boat. Just after her leap, how fast is she moving relative to the water? Express your answer with the appropriate units Value Units Submit Request Answer ▼ Part B Just after her leap, how fast is the canoe moving relative to the water? Express your answer with the appropriate units. (c)EValue Units
The speed of Riley relative to the water is 1.7 m/s. and the speed of canoe relative to the water is 0 m/s.
How fast is Riley moving relative to the water?The equation needed to solve the problem is the following:
Final Velocity = Initial Velocity + (Acceleration × Time)
The steps to solve for speed of Riley are the following:
Mass of Riley = 62 kg
Mass of canoe = 21 kg
Speed of leap relative to the boat = 1.7 m/s
By using the equation for conservation of momentum (also known as the center of mass formula):
m₁v₁ + m₂v₂ = (m₁ + m₂)vf
Solve for the unknown variable: vf = (m₁v₁ + m₂v₂) / (m₁ + m₂)
Plugging in the values given, you get: vf = (62 kg × 1.7 m/s) / (62 kg + 21 kg) = 1.2 m/s
Therefore, Riley is moving at 1.2 m/s relative to the water.
Velocity of the canoe relative to the water can be determined by using the equation for conservation of momentum (also known as the center of mass formula):
m₁v₁ + m₂v₂ = (m₁ + m₂)vf
v₂ = [(m₁ + m₂)vf - m₁v₂] / m₂
Plugging in the values given, you get: v₂ = [(62 kg + 21 kg) × 1.2 m/s - 62 kg × 1.7 m/s] / 21 kg = 0 m/s
Therefore, the canoe is not moving relative to the water.
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the sun is shining on james and has created a shadow that is 8 feet long what is the distance from his head to his shadow
The distance from James's head to his 8-foot long shadow is 8 feet.
To find this, we can use the simple triangle relationship: opposite side (the shadow) is equal to the hypotenuse (the distance between James's head and his shadow) times the cosine of the angle formed by the sun's rays.
The distance from James's head to his shadow can be calculated using trigonometry. The tangent function can be used for this purpose.
To calculate the distance from James's head to his shadow, follow these steps:
Firstly, draw a right triangle with one of its angles adjacent to James's head and the other adjacent to the base of the shadow. The hypotenuse of the triangle is the line between James's head and the tip of the shadow.
Let x be the distance from James's head to the base of the shadow. The hypotenuse is the square root of (x^2 + 8^2).
Use the tangent function to find the value of x. tan(angle) = opposite/adjacent. In this case, the angle is the angle of elevation of the sun, which can be determined from the time of day and the location.
If the angle is not known, it can be assumed to be 45 degrees. tan(45) = opposite/adjacent.
The opposite side is 8, so: x = 8/tan(45) = 8/1 = 8 feet.
Therefore, the distance from James's head to his shadow is 8 feet.
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A ball of mass, m is thrown straight up and rises h after leaving your hand, it momentarily stops. Acceleration due to gravity g is downward. Ignore air resistance. Part A (5 points): If the ball is the system, and the Earth is the surroundings, what is the change in potential energy, ΔUsys of the system, and what is the work done, Wsurr by the surroundings ? Δ Usys = 0; Wsurr = -mgh Δ Usys = mgh; Wsurr = -mgh Δ Usys = mgh; Wsurr = 0 Δ Usys = 0; Wsurr = 0 Δ Usys = -mgh; Wsurr = 0
the correct answer is option (C). If the ball is the system, and the Earth is the surroundings, the change in potential energy, ΔUsys of the system is mgh, and the work done, Wsurr by the surroundings is 0. Therefore.
In this case, the ball is hurled straight up, reaching a height of h before briefly coming to a stop. We assume the ball to be the system and the Earth to be the surroundings in order to calculate the change in potential energy of the system and the work performed by the surroundings. The system's change in potential energy is mgh because the ball's gravitational potential energy grows as it ascends to a height of h. Yet, because the environment is not subjected to any labour from the ball during its rise and descent, there is no work done by the environment.
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Using the heat of vaporization of benzene, 395 J/g, calculate the grams of benzene that will condense at its boiling point if 8.44 kJ is removed.
Considering the heat of vaporization of benzene, the mass that will evaporate, at the boiling point, if 8.44 kJ/g of heat is extracted is 21.36 g.
Given the heat of vaporization of benzene, 395 J/g and the heat removed, 8.44 kJ, we can determine the mass of benzene that condenses by converting the heat removed to J/g as follows:
Qv = 8.44 kJ/g · 1000 J / 1 kJ = 8440 J/g
Hence, mass of benzene that condenses can be found by dividing the heat removed by the heat of vaporization as shown:
mass = heat removed / heat of vaporization
m = 8440 J/g / 395 J/g
m = 21.36 g
Therefore, 21.39 g of benzene will condense at its boiling point if 8.44 kJ is removed.
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The temperature of a gas stream is to be measured by a thermocouple whose junction can beapproximated as a 1.2-mm-diameter sphere. The properties of the junction are k =35 W/m °C, p=8500kg/m3, and Cp = 320 J/kg °C, and the heat transfer coefficient between the junction and the gas is h=65W/m2 °C. Determine how long it will take for the thermocouple to read 99 percent of the initialtemperature difference. (∅/∅i= 0.01)
it will take 30.65 minutes for the thermocouple to read 99 percent of the initial temperature difference. (∅/∅i = 0.01).
The temperature of a gas stream is to be measured by a thermocouple whose junction can be approximated as a 1.2-mm-diameter sphere. So, the radius, r = 0.6 mm = 0.0006 m, the volume of the sphere, V = (4/3)πr³, and the area of the sphere, A = 4πr².
The properties of the junction are k = 35 W/m °C, p = 8500 kg/m³, and Cp = 320 J/kg °C, and the heat transfer coefficient between the junction and the gas is h = 65 W/m² °C.
We have, thermal conductivity of the sphere = k = 35 W/m °C, density of the sphere = p = 8500 kg/m³, specific heat of the sphere = Cp = 320 J/kg °C, and heat transfer coefficient between the sphere and the gas, h = 65 W/m² °C.
The initial temperature difference is given by, ΔT₀ = 1°C = 1 K. Let, the time taken for the thermocouple to read 99% of the initial temperature difference, ΔT99 = 0.99 K.
Let, the thermal diffusivity of the sphere be,
α = k / (pCp) = (35 W/m °C) / (8500 kg/m³ x 320 J/kg °C) = 0.000012868 m²/s.
And, the Biot number is given by, Bi = (h x A) / k = [(65 W/m² °C) x 4π(0.0006 m)²] / (35 W/m °C) = 0.0492.
The equation for the unsteady-state temperature profile of a sphere is, θ(r,t) = Σ [(-1)n+1 / n] exp(-n²π²αt / r²) sin (nπr / R), where R is the radius of the sphere. We can estimate the time taken for the thermocouple to read 99% of the initial temperature difference using a semi-log plot of θ/ΔT vs. t/ti.
This plot is linear and of the form, θ/ΔT = 1 - A exp (-Bt/ti), where A = 0.01 and B = (nπ/R)².So, θ/ΔT = 0.99 = 1 - A exp (-Bt/ti), or 0.01 exp (-Bt/ti) = 0.01/0.99, or exp (-Bt/ti) = 1/99, or -Bt/ti = ln (1/99), or t/ti = ln (99).
Therefore, the time taken for the thermocouple to read 99% of the initial temperature difference is, ti = t / ln (99) = (0.000012868 m²/s) (0.6 mm)² / (35 W/m °C) ln (99) = 1838.98 s or 30.65 minutes.
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Sam (85 kg) takes off up a 50-m-high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 220 N. He keeps his skis tilted at 10 degree after becoming airborne. How far does Sam land from the base of the cliff?
Sam (85 kg) takes off up a 50-m-high, 10 degree frictionless slope on his jet-powered skis. The skis have a thrust of 220 N. He keeps his skis tilted at 10 degree after becoming airborne. Sam lands about 109.9 meters from the base of the cliff.
To solve this problem, we can use the conservation of energy principle. At the bottom of the slope, all of Sam's energy is in the form of potential energy:
Potential energy = mgh
where m is Sam's mass (85 kg), g is the acceleration due to gravity [tex](9.81 m/s^2)[/tex], and h is the height of the slope (50 m).
Potential energy = [tex](85 kg) \times (9.81 m/s^2) \times (50 m) = 41,287.5 J[/tex]
As Sam takes off up the slope, his potential energy is converted to kinetic energy and then to a combination of kinetic and potential energy as he becomes airborne. We can use the conservation of energy to find Sam's speed at the top of the slope:
Potential energy at bottom = Kinetic energy at top
[tex]mgh = (1/2)mv^2[/tex]
where v is Sam's speed at the top of the slope.
[tex]v = \sqrt{(2gh)} = \sqrt{(2 \times 9.81 m/s^2 \times 50 m)} = 31.3 m/s[/tex]
Now, we can use Sam's speed and the angle of his skis to find his horizontal velocity:
Horizontal velocity = v cos(theta)
where theta is the angle of the skis after becoming airborne (10 degrees).
Horizontal velocity = 31.3 m/s x cos(10 degrees) = 30.2 m/s
Finally, we can use the horizontal velocity and Sam's hang time to find the distance he travels:
Distance = Horizontal velocity x Hang time
where hang time is the time Sam spends in the air. Hang time can be found using the formula:
Hang time = (2v sin(theta)) / g
Hang time = (2 x 31.3 m/s x sin(10 degrees)) / 9.81 [tex]m/s^2[/tex] = 3.64 s
Distance = 30.2 m/s x 3.64 s = 109.9 m
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Light of 630 nm wavelength illuminates two slits that are 0.25 mm apart. FIGURE EX33.5 shows the intensity pattern seen on a screen behind the slits. What is the distance to the screen?
The distance to the screen from the two slits is 4.0 meters
Distance is the total distance traveled by an object over a specific time interval.
The distance can be calculated using the equation d = λ/(2a), where
d is the distance to the screen, λ is the wavelength of the light (630 nm in this case), and a is the separation of the two slits (0.25 mm in this case).Plugging these values in, we get: d = 630 nm / (2 * 0.25 mm) = 4.0 m. The distance to the screen from the two slits is 4.0 meters, as seen in Figure EX33.5.
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what happens to the domains of an object when it becomes magnetized?
When an object becomes magnetized, the domains within the object align themselves in a particular direction. A domain is a small region within a magnetic material that behaves like a tiny magnet, with its own north and south poles.
In an unmagnetized object, these domains are randomly oriented, so their magnetic fields cancel each other out and the material as a whole has no net magnetic field. When an external magnetic field is applied to the material, the magnetic domains become aligned with the external field, resulting in a net magnetic field for the material. If the external field is strong enough, it can overcome the random orientation of the domains and cause them to align in the direction of the external field. Once the domains are aligned, the object will retain its magnetism even after the external magnetic field is removed. This is because the domains will remain in their aligned state until another external magnetic field is applied to the object and causes them to realign.
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Alice holds a small battery operated device used for tuning instruments that emits the frequency of middle C (262 Hz) while walking with a constant speed of 4.68 m/s toward a building which presents a hard smooth surface and hence reflects sound well. (Use343 m/s as the speed of sound in air.)
(a) Determine the beat frequency Alice observes between the device and its echo. (Enter your answer to at least 1 decimal place.)
(b) Determine how fast Alice must walk away from the building in order to observe a beat frequency of 6.19 Hz.
(A) Alice observes a beat frequency of approximately 3.9 Hz between the device and its echo. (B) Alice must walk away from the building at a speed of approximately 7.05 m/s to observe a beat frequency of 6.19 Hz.
(A) The given values are:
Speed of Alice, vA = 4.68 m/s.
The frequency emitted by the device, f1 = 262 Hz
Speed of sound in air, v = 343 m/s(a)
The beat frequency, f beat is given by the formula: fbeat = |f1 - f2| where f2 is the frequency of the reflected sound.
Since the speed of sound is reflected, the distance traveled by the sound to the building and back is 2d.
Therefore, the time taken is given by t = 2d/v.
The frequency f2 is given by f2 = v/(2d).
The distance d = vt/2 = (vA t)/2
The time t is given by: t = d/vA
The frequency f2 is given by f2 = v/(2d) = vA/(2v t)
Therefore, the beat frequency is: fbeat = |f1 - f2| = |262 - vA/(2v t)|
Thus, substituting the given values, we get: fbeat = |262 - 343/(2 × 4.68 × t)|
To solve this, we can use trial and error method.
We can check if fbeat is approximately equal to 2, 3, 4, 5, or 6 Hz.
Using t = 0.01 s, we get: fbeat = |262 - 343/(2 × 4.68 × 0.01)|≈ 4.4 Hz
Using t = 0.011 s, we get: fbeat = |262 - 343/(2 × 4.68 × 0.011)|≈ 3.9 Hz
Therefore, Alice observes a beat frequency of approximately 3.9 Hz between the device and its echo.
(b) Let's suppose that Alice walks with a velocity of vA' away from the building. Therefore, the distance traveled by the sound in the same time interval t = d/vA' is d' = vA' t/2.The time taken is given by t = d/vA = d'/vA'
Now, the frequency f2 is given by f2 = v/(2d') = vA'/(2v t)
The beat frequency is:fbeat = |f1 - f2| = |262 - vA'/(2v t)|
Thus, substituting the given values, we get: fbeat = |262 - 343/(2 × vA' × t)|
Let's suppose that fbeat = 6.19 Hz.
Using trial and error, we get that t ≈ 0.018 s.
Substituting this value, we get:6.19 = |262 - 343/(2 × vA' × 0.018)|
Therefore, vA' ≈ 7.05 m/s
Thus, Alice must walk away from the building at a speed of approximately 7.05 m/s to observe a beat frequency of 6.19 Hz.
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what device is used to shunt transient current to ground in the event of an indirect lightning strike?
In the event of an indirect lightning strike, a Surge Protection Device (SPD) is used for shunting transient current to the ground. An SPD is a protective device that limits the voltage supplied to an electrical system by either blocking or shorting to ground any unwanted voltages above a safe threshold. This can help protect against damage from transient current, a short, high-energy burst of electricity.
A surge protector is an electrical device that protects electronic devices from power surges and other electrical disturbances. The device will shield the equipment that is plugged into it from the spikes that are present in an electrical supply.The term “surge protector” is frequently used in reference to a category of products that is also known as a “transient voltage suppressor.” This name provides insight into how these devices work. They suppress transient voltage, which is a sudden surge of voltage that is brief in nature
.How do surge protectors work?
Surge protectors work by preventing transient voltage spikes from reaching sensitive electrical equipment. These devices typically consist of a metal oxide varistor, which is a component that is used to divert any unwanted voltage away from sensitive electronics and toward a grounded element.The varistor is connected to a metal oxide varistor, which is responsible for conducting the unwanted voltage away from the equipment and toward the ground. Surge protectors will reduce voltage to a safe level by grounding the unwanted voltage. Surge protectors are used to protecting a wide range of electronic devices, including computers, audio equipment, and video equipment.
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Transient current refers to an electrical current that flows for a brief period. Transient currents are caused by temporary changes in voltage, such as those caused by electrical discharges, power outages, and other events. Surge currents are another name for transient currents, and they are often used interchangeably.
A lightning strike is an electrical discharge from the atmosphere to the earth's surface. Thunderstorms, which are associated with lightning, are the most frequent natural cause of the electrical discharge. A lightning bolt can produce extremely high voltages and currents, posing a significant threat to electrical systems and the people who operate them.
A surge protector is a device that is intended to protect electrical devices from voltage spikes, surges, and other power fluctuations. Surge protectors work by shunting transient currents to the ground in the event of an indirect lightning strike. They can also be used to safeguard against other types of power surges, such as those caused by power outages, grid switching, and other issues. Surge protectors are often utilized in industrial and commercial settings, as well as in homes.
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