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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The straight section of the line in figure 10 can be used to calculate the useful power output of the kettle explain how
Using the line's straight segment in figure 10, it is possible to determine the usable power output of the kettle.
The period that the kettle is heating the water up until it reaches boiling point is depicted by the straight segment of the line in figure 10. Both the power input to the kettle and the rate of energy transfer to the water remain constant throughout this period. Hence, by dividing the energy that was transmitted to the water during this period by the whole amount of time, the usable power output of the kettle can be determined. The straight section's slope, which reflects the rate of energy transfer, and horizontal distance, which indicates the elapsed time, may be used to calculate this. The energy transmitted is calculated by dividing the rate of energy transmission by the amount of time.
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How much force is required to accelerate a 5kg mass at 20m/s 2 ?
Нам не дано коэффициент трения, значит, можно не учесть силу трения. От этого, по второму закону Ньютона, F=ma=5×20=100 Н.
И это всё!
Imagine sitting on a merry-go-round and riding along as it spins. Assuming you are not grabbing it anywhere and are not moving with respect to the platform,
A. static friction (directed inwards) causes you to accelerate.
B. you are not accelerating because you aren't moving on the platform.
C. static friction (directed outwards) causes you to accelerate.
D. sliding friction makes you accelerate inwards.
The correct option is: Static friction (directed outwards) causes you to accelerate. (Option C)
When you sit on a merry-go-round, you are not moving relative to the platform. Therefore, you are not in motion in respect to the reference frame of the platform.
The question is asking you to determine the force that causes you to accelerate as the merry-go-round spins.
Static friction is the force that keeps an object at rest or keeps it moving in a straight line when a force is applied to it.
When you're riding a merry-go-round and it starts to spin, static friction force helps you move outwards. This force opposes the force that pulls you towards the center of the platform, i.e., centripetal force.
So the correct option is C: Static friction (directed outwards) causes you to accelerate.
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An object starts at rest in position A on the track shown, then slides to position B. Friction acts on the object over the entire track. Which equation can you use to find the object's velocity at position B?
Question 7 options:
- mgy3 + Wfriction = mgy2
- mgy2 + Wfriction = (1/2)mv2 + mgy1
- mgy3 + Wfriction = (1/2)mv2
- mgy3 + Wfriction = (1/2)mv2 + mgy2
- Wfriction = (1/2)mv2 + mgy3 + mgy2
- mgy3 = Wfriction + (1/2)mv2 - mgy2
- mg(y3 - y2) = (1/2)mv2
- Wfriction = (1/2)mv2 + mgy2
The equation that can be used to find the object's velocity at position B is [tex]mgy_3 + W_{friction} = (1/2)mv^2 + mgy_2[/tex].
What is friction?Friction is the resistance encountered when one object moves over another. Friction opposes the movement of objects and is dependent on the roughness of the surfaces, the force pressing the objects together, and the surface area. It is a force that opposes movement, and it occurs when two surfaces come into touch. It operates in the opposite direction to movement and is always parallel to the surface of contact.
What is Velocity?Velocity is a measure of the displacement of an object per unit time in a given direction. The distance traveled by an object in a specific time period and in a specific direction is referred to as displacement.
As a result, velocity is a vector quantity because it has both magnitude and direction. It is calculated by dividing the displacement by the time taken, according to the definition.
Since friction is acting over the entire track, this equation takes into account the work done by friction to reduce the object's velocity from its initial value of 0 m/s at position A to its final velocity at position B.
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A ball rolls along a horizontal track in a certain time. If the track has a small upward dent in it, the time to roll the length of the track will be:
a. less
b. more
c. the same
Explanation:
More....it will have to travel a greater length to go up and over the dent, so it will take longer
a 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s. what is the angle of the pendulum?
A 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s then the angle of pendulum is 14.68°.
Given:
Mass of the object = 0.4kg
Length of string = 0.9m
Period of conical pendulum = 1.4s
The angle of pendulum is calculated by using this formula :
T = 2π(r/g)1/2
where, T is the time period of the circular motion g is acceleration due to gravity r is radius of the circle
Let us assume, Angle made by the string with the vertical axis = αNow, Radius of circle can be given as,
R = l.sinα
Given the period of the conical pendulum as 1.4s
we can find the acceleration due to gravity as follows = 2π(r/g)1/2r = l.sinα2π(r/g)1/2 = Tg = 4π2(l.sinα)2/T2g = 4π2(l2sin2α)/T2sinα = gT2/4π2l2Sinα = (9.8 m/s2× 1.4 s2)/(4π2 × (0.9 m)2)Sinα = 0.253α = sin-1(0.253)α = 14.68°
Hence, the angle made by the string with the vertical axis is 14.68°.
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when you look at a spiral that appears to move inward for about a minute, and then look at a stationary object, the object will briefly appear to ......
When you look at a spiral that appears to move inward for about a minute, and then look at a stationary object, the object will briefly appear to move outwards. This phenomenon is known as the motion aftereffect (MAE).
After staring at the spiral for about a minute, your brain becomes accustomed to the constant motion of the spiral. When you look away from the spiral and fix your gaze on a stationary object, your brain continues to perceive motion in the opposite direction (outwards).
This is why the stationary object appears to move outwards for a brief period. The motion aftereffect is an example of the adaptation process that takes place in the visual system. It is a perceptual illusion that occurs when the brain is exposed to a particular type of visual stimulus for a prolonged period of time.
Hence, when you look at a spiral that appears to move inward for about a minute, and then look at a stationary object, the object will briefly appear to move outwards.
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Taking the following list on an item-by-item basis (i.e., without considering the other listed factors), a maintenance expenditure should be capitalized if the expenditure:
increases the salvage value of the asset.
extends the useful life of the asset.
A maintenance expenditure should be capitalized if it increases the salvage value of the asset or extends the useful life of the asset.
An expenditure is a payment made in return for a product or service. Capital expenditure is money spent by a company on long-term assets like equipment and buildings.
Capitalizing refers to recording a cost or expense on the balance sheet for a future period rather than recognizing it immediately in the current period.
Capitalizing expenditure means the company will recognize the expenditure as an asset, which will be amortized over its useful life as opposed to expenses in the current period.
Therefore, a maintenance expenditure should be capitalized if the expenditure increases the extends the useful life of the asset.
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determine whether each geologic feature is being caused by tensional, compressional, or shear stresses by analyzing the directions of the forces being applied.
In any case, the type of force that is responsible for creating a particular geological feature depends on the direction and magnitude of the forces that are acting on it.
Geological features are landforms that are made up of natural formations. A wide variety of geological features exist in nature, including mountains, valleys, canyons, caves, and others.
There are a variety of geological features that can be created as a result of tensional, compressional, or shear stresses.
Let's take a closer look at each type of stress:
Tensional: Tensional forces act to pull rocks apart. This can result in the formation of fault-block mountains, valleys, and rifts.
Compressional: Compressional forces act to push rocks together. This can lead to the creation of mountain ranges, folded mountains, and plateaus.
Shear Stresses: Shear stresses act to twist or bend rocks. This can result in the formation of faults, folds, and other geological features.
The forces that create geological features are typically produced by the movement of tectonic plates beneath the earth's surface.
When two tectonic plates come together, they can create compressional forces. When they move apart, they can create tensional forces.
When they slide past each other, they can create shear stresses. Other forces can also play a role, such as erosion or the buildup of sediment over time.
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A horizontal force of magnitude 35.0N pushes a block of mass 4.00kg across a floor where the coefficient of kinetic friction is 0.600. (a) how much work is done by the applied force on the block-floor system when the block slides through a displacement of 3.00m across the floor? (b) during that displacement the thermal energy if the block increases by 40.0J. what is the increase in thermal energy of the floor? (c) what is the increase in the kinetic energy of the block?
Answer to following (a) , (b) and (c) question are: 63.00 J, 40.0 J, 63.00 J
(a) The work done by the applied force on the block-floor system when the block slides through a displacement of 3.00m across the floor can be calculated by multiplying the applied force (35.0 N) and the displacement (3.00 m), with a coefficient of kinetic friction (0.600) for the system. Thus, the work done is 35.0N * 3.00m * 0.600 = 63.00 J.
(b) The increase in the thermal energy of the floor during the displacement of 3.00m is equal to the thermal energy of the block (40.0 J), since the total thermal energy of the block-floor system remains constant. Therefore, the increase in thermal energy of the floor is 40.0 J.
(c) The increase in the kinetic energy of the block is equal to the work done by the applied force, i.e., 63.00 J.
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suppose a car approaches a hill and has an initial speed of 102 km/h at the bottom of the hill. the driver takes her foot off of the gas pedal and allows the car to coast up the hill.
If the car has the initial speed stated at a height of h = 0, how high, in meters, can the car coast up a hill if work done by friction is negligible?
The initial speed of the car that approaches a hill is 102 km/h. The driver takes her foot off of the gas pedal and allows the car to coast up the hill. If the car has the initial speed stated at a height of h = 0, the height the car can coast up a hill is 34.3 meters if work done by friction is negligible.
What is Work done?Initial Energy = Potential Energy
Hence, the Potential Energy formula is given as:
PE = mgh
where, PE = Potential Energy (Joules)
mg = mass × gravity
h = height
Potential Energy at h = 0 is given as follows:
PE₀ = mgh₀
PE₀ = 0mg
PE₀ = 0
Potential Energy at h = 1 is given as follows:
PE₁ = mgh₁
Let's equate the two potential energies and solve for h₁:
PE₁ = PE₀ (since work done by friction is negligible)
mgh₁ = 0h₁ = 0
Therefore the height of the car that can coast up a hill is 34.3 meters if work done by friction is negligible.
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Physics Help Requested Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g=30 m/s2. When he releases the ball from chin height without giving it a push, how will the ball's behavior differ from its behavior on Earth? Ignore friction and air resistance. (Select all that apply.)a. It will take more time to return to the point from which it was released.b. It will smash his face. Its mass will be greater.c. It will take less time to return to the point from which it was released. d, It will stop well short of his face.
On a planet with more massive gravity, such as [tex]g = 30 \ m/s^2[/tex], the ball released from chin height will take less time to return to the point from which it was released, due to the increased acceleration due to gravity.
It will take less time to return to the point from which it was released. The acceleration due to gravity is much stronger on this planet, so the ball will accelerate faster as it falls toward the ground. This means that it will reach its lowest point more quickly and then rise back up to its starting point more quickly as well.
Also, the mass of the ball is not affected by the strength of the gravitational acceleration on the planet.
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Three identical conducting spheres are charged as follows. Sphere A is positively charged, sphere B is negatively charged with a different magnitude of net charge than that of sphere A, and sphere C is uncharged. Spheres A and B are momentarily touched together and separated, then spheres B and C are briefly touched together and separated. After that series of processes is completed, which of the following interactions, if any, can be used as evidence to determine whether sphere A or sphere B had the initially larger magnitude of charge? A Sphere C is repelled from sphere A. B Sphere C is repelled from sphere B. Sphere A is repelled from sphere B. D It cannot be determined from observing whether the spheres repel, because they all have the same sign of charge.
The answer is C. Sphere A is repelled from sphere B
Step by step explanation:
The question is asking which of the interactions between sphere A, B, and C can be used as evidence to determine which one had the initially larger magnitude of charge. This is because if sphere A has a larger magnitude of charge than sphere B, then when spheres A and B are touched and separated, the charge of sphere A would be transferred to sphere B, causing a conduction of charge.
This means that after the processes are completed, the charge of sphere A and B will have reversed - meaning that sphere A will now have the same, but opposite sign of charge as sphere B. As a result, when sphere A and B are close to each other, their charges will repel, so Sphere A is repelled from sphere B.
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you have an rc circuit with a time constant of 5.35 s. if the total resistance in the circuit is 231.2 k , what is the capacitance of the circuit (in f)? don't type the units into the answer box.
The capacitance of the circuit (in f) is 2.31×10⁻⁵F for the rc circuit with a time constant of 5.35 s. if the total resistance in the circuit is 231.2 k.
What is the capacitance of the circuit?The capacitance of an RC circuit can be calculated using the equation C = τ/(R), where τ is the time constant, R is the total resistance, and C is the capacitance. For this RC circuit, the time constant is 5.35s and the total resistance is 231.2 k. Therefore, the capacitance is 5.35s/(231.2k) = 2.31×10⁻⁵F.
Time constant of the RC circuit, τ = 5.35s
Total resistance in the circuit, R = 231.2 kΩ = 231200 Ω
Capacitance of the circuit = ?
We know that, Time constant (τ) of a RC circuit = R × C.
where, R is the resistance in ohms, C is the capacitance in farads. Substitute the given values in the above equation:
τ = RC
5.35 s = R × C231200 Ω × C = 5.35 s
C = 5.35 s / 231200 Ω
C = 2.31 × 10⁻⁸ F.
Therefore, the capacitance of the circuit is 2.31 × 10⁻⁸ F.
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A 200 g air-track glider is attached to a spring. The glider is pushed 10.0 cm against the spring, then released. A student with a stopwatch finds that 10 oscillations take 12.0 s. What is the spring constant? A 200 g ball is tied to a string. It is pulled to an angle of 8.00degree and released to swing as a pendulum. A student with a stopwatch finds that 10 oscillations take 12.0 s. How long is the string?
The spring constant of spring is 39.9 N/m and the length of the string is about 47.5 meters.
What is spring constant?Mass of the air-track glider (m) = 200 g = 0.2 kg
Displacement of the air-track glider (x) = 10.0 cm = 0.1 m
Number of oscillations (n) = 10
Time taken for 10 oscillations (t) = 12.0 s
T = 2π√(m/k)
where, T is the time period of oscillation. Substituting the given values, we get:
12 s = 2π√(0.2 / k)
Solving for k, we get:
The spring constant is 39.9 N/m.
Mass of the ball (m) = 200 g = 0.2 kg
Angle of displacement (θ) = 8.00°
Number of oscillations (n) = 10
Time taken for 10 oscillations (t) = 12.0 s
T = 2π√(L/g)
where, T is the time period of oscillation and g is the acceleration due to gravity. Substituting the given values, we get:
12 s = 2π√(L/9.8)
Solving for L, we get:
L = (12/2π)² × 9.8 = 47.5 m
Therefore, the length of the string is 47.5 meters.
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A solar sailplane is going from Earth to Mars. Its sail is oriented to give a solar radiation force of FRad = 7.70 × 102 N. The gravitational force due to the Sun is 173 N and the gravitational force due to Earth is 1.00 × 102 N. All forces are in the plane formed by Earth, Sun, and sailplane. The mass of the sailplane is 14,900 kg. What is the magnitude of the acceleration on the sailplane? Answer in m/s2
The sailplane which is going from Earth to Mars is accelerating at 0.033 m/s² in the direction of solar radiation force.
The force of gravity is a force that arises as a consequence of the mutual attraction of two objects. This gravitational force is usually exerted between two physical objects. Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is proportional to the product of their masses.
Acceleration is the rate at which an object changes its speed or direction. Acceleration is a vector quantity that can be positive or negative. If the acceleration is negative, the object slows down. If the acceleration is positive, the object speeds up.
The acceleration on the sailplane can be determined using the following formula:
[tex]F_{net} = ma[/tex]
Where Fnet is the net force acting on the sailplane, m is the mass of the sailplane a is the acceleration on the sailplane.[tex]F_{net} = ma[/tex]
The net force acting on the sailplane can be calculated as:
[tex]F_{net} = F_{rad} - F_{gravitySun} - F_{gravityEarth}[/tex]
Where [tex]F_{rad}[/tex] is the solar radiation force, [tex]F_{gravitySun}[/tex] is the gravitational force due to the sun, and [tex]F_{gravityEarth}[/tex] is the gravitational force due to Earth.
Putting the given values in the above formula:
[tex]F_{net} = 7.70 \times 10^2 N - 173 N - 1.00 \times 10^2 N = 497 N[/tex]
The acceleration on the sailplane is given as:
[tex]a = F_{net} / ma = (497\ N) / 14,900 \ kg = 0.033 \ m/s^2[/tex]
The magnitude of the acceleration on the sailplane is 0.033 m/s² (rounded to three significant figures).
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When a ball bounces against a wall there will be large change in velocity in short period of time. This means the ____ is large, hence the net ___ must be proportionately large as well.
A change in velocity in short period of time means the acceleration is large, hence the net force must be proportionately large as well.
What is a force?A force is a physical quantity that induces a body to undergo an alteration in speed, direction of motion, or shape. A force can be classified as a push or a pull. When forces are equal, the forces are balanced and the object is not moving. Otherwise, if the forces are not equal, making it unbalanced will not give the object any movement.
The force that induces the change in the speed or direction of an object is referred to as a net force. The net force is equal to the product of the mass of the object and its acceleration. Newton (N) is the unit of measurement for force.
When a ball bounces against a wall, there will be a large change in velocity in a short period of time. This means the acceleration is large, hence the net force must be proportionately large as well.
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at what angle above the horizon is the sun when light reflecting off a smooth lake is polarized most strongly?
The sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.
When unpolarized light reflects off a smooth surface, such as a lake, it becomes polarized in a direction perpendicular to the surface. The angle at which this polarization is strongest is known as the Brewster angle, and can be calculated using the formula:
θB = arctan(n2/n1)
where θB is the Brewster angle, n1 is the index of refraction of the medium the light is coming from, and n2 is the index of refraction of the medium the light is entering.
For water, the index of refraction is approximately 1.33, and for air it is approximately 1.00. Plugging these values into the formula, we get:
θB = arctan(1.33/1.00) = 53.1 degrees
However, this is the angle at which the light is reflected off the surface in a direction perpendicular to the surface. To find the angle above the horizon at which the light is polarized most strongly, we need to subtract 90 degrees from the Brewster angle:
37 degrees = 90 degrees - 53.1 degrees
Therefore, the sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.
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Acceleration due to gravity is 9.8 m/s2 on the surface of Earth, and at orbits 200 miles above the surface of Earth, where the space shuttle orbits, the acceleration is
Acceleration due to gravity is 9.8 m/s2 on the surface of Earth, and at orbits 200 miles above the surface of Earth, where the space shuttle orbits, the acceleration is 8.78 m/s².
What is gravitational force?The reason for this difference in acceleration is that the gravitational force on an object is inversely proportional to the square of the distance between them.
Thus, the further an object is from the Earth's surface, the weaker the gravitational force acting on it. This is why objects in orbit around the Earth experience less acceleration due to gravity than objects on the surface of the Earth.
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What two planets are coming together?
The two planets that are coming together are Saturn and Jupiter. On December 21st, 2020, the two planets will be at their closest point, an event known as the Great Conjunction.
To observe the Great Conjunction, look in the direction of the southwest sky shortly after sunset. The two planets will appear to be close together and will look like one bright star. Make sure to look for them with binoculars or a telescope if you can, as you'll get a better view.The Great Conjunction occurs because Saturn and Jupiter have different orbital periods. Jupiter completes its orbit around the Sun every 11.86 Earth years, while Saturn takes 29.5 Earth years. This means that their orbits don't intersect and they don't come this close together very often. The next time the two planets will come this close together will be in 2080, so be sure to take advantage of this rare opportunity to witness this event in 2020.For more questions on Great Conjunction
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member bc exerts on member ac a force p directed along line bc. knowing that p must have a 325-n horizontal component, determine (a) the magnitude of the force p, (b) its vertical component.
(a) The magnitude of the force p=325 / cos θPart, (b) Vertical component is 325 tanθ
(a) Given: Force F = P And horizontal component Fcos θ = 325N. Here, θ is the angle made by the force with the horizontal, and θ is unknown. According to the figure, member AC is inclined at an angle θ to the horizontal.
Let's resolve the force P into vertical and horizontal components. So, vertical component Fsine θ and horizontal component Fcos θ, where θ is the angle made by the force with the horizontal, and θ is unknown.
Thus, we get: Fcos θ = 325Fcos θ / F = 325 / cos θPart
(b) Vertical component = Fsine θ = (F)(sinθ)Vertical component = (325 / cosθ)(sinθ) = 325 tanθ
Thus, the magnitude of the force p is 325 / cosθ, and the vertical component of the force is 325 tanθ.
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a big block of mass m(10kg) slides down a frictionless inclined at an angle 30 with the horizontal table. initially the block is at the top of the incline at rest. determine the speed of the block at the bottom of the incline
When the big block of mass m(10kg) slides down a frictionless inclined at an angle 30 with the horizontal table, the speed of the block at the bottom of the incline is 3.14 m/s.
Given that
Mass of the block, m = 10 kg.
Angle of inclination, θ = 30°
Initial velocity, u = 0.
Frictional force, f = 0.
Using the formula for gravitational force, F = mg
where, g = 9.8 m/s² (acceleration due to gravity)
F = mg= 10 kg × 9.8 m/s²= 98 N
The component of gravitational force that acts parallel to the incline, Fsinθ is responsible for the acceleration of the block. Fsinθ = ma; Where a is the acceleration of the block.
a= (98 N)sin 30° / 10 kg= 4.9 m/s²
Using the formula for speed, v = u + at where,
u = initial velocity = 0m/s
t = time taken = time taken to slide from top to bottom of the incline.= √(2h/g) where,
h = height of the incline = 2 m (since the mass is at rest initially at the top of the incline).
Therefore, t = √(2 × 2 m / 9.8 m/s²)= 0.64 s
Substituting the values in the above formula, v = u + at= 0 + (4.9 m/s² × 0.64 s)= 3.14 m/s.
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Artificial gravity. One way to create artificial gravity in a space station is to spin it. Part A If a cylindrical space station 325 m in diameter is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to g ? f = nothing rpm
The space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.
Part A:If a cylindrical space station with a diameter of 325 m is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to g?The acceleration of the outermost points is given as g. To create artificial gravity, the space station must spin about its central axis. To determine the required rpm, use the formula for acceleration due to centripetal force, which is given by:a = rω2Where, a is the acceleration due to centripetal force, r is the radius of the circle, and ω is the angular velocity of the object in radians per second. One full rotation equals 2π radians. Therefore, the angular velocity can be computed asω = 2πnwhere n is the number of revolutions per second. To transform it to rpm, use the formula:n = (r.p.m)/(60s)Substitute the values in the formula to obtain the solution as follows:g = a = rω2r = 325/2 = 162.5ma = g = 9.8 m/s2ω = 2πn⇒ω2 = (2πn)2⇒ω2 = 4π2n2Substitute the values in the formula for a to obtain:rω2 = g⇒(162.5 m)(4π2n2) = 9.8 m/s2n = 1.49 rpmTherefore, the space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.
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A kangaroo is capable of jumping to a height of 2.62m. Determine the takeoff speed of the kangaroo.
Answer: 7.17
Explanation:
Maximum height reached by Kangaroo H=2.62
Final velocity at the maximum height v=0
Acceleration due to gravity g=−9.8 m/s2
Using v2−u2=2gH∴ 0−u2=2(−9.8)(2.62)
⟹ u=2(9.8)(2.62)=7.17 m/s
A resistor is constructed by shaping a material of resistivity p into a hollow cylinder of length L and with inner and outer radii ra and rb, respectively (Fig. P27.66). In use, the application of a potential difference between the ends of the cylinder produces a current parallel to the axis, (a) Find a general expression for the resistance of such a device in terms of L, p, ra, and rb. (b) Obtain a numerical value for. R when L = 4.00 cm, ra = 0.500 cm, rb = 1.20 cm, and p = 3.50 times 105 Ohm m. (c) Now suppose that the potential difference is applied between the inner and outer surfaces so that the resulting current flows radially outward. Find a general expression for the resistance of the device in terms of L, p, Figure P27.66 ra, and rb. (d) Calculate the value of R, using the parameter values given in part (b).
Explanation:
Refer to pic...........
Hooke's law: Consider a plot of the displacement (x) as a function of the applied force (F) for an ideal elastic spring. The slope of the curve would be A) the mass of the object attached to the spring. B) the reciprocal of the acceleration of gravity. C) the spring constant. D) the acceleration due to gravity. E) the reciprocal of the spring constant.
Hooke's law: the slope of the curve would be the spring constant (C).
What is Hooke's law?Hooke's law is a principle of physics which states that the force F needed to extend or compress a spring by some distance x scales linearly with respect to that distance.
F = kx
where k is the spring constant and x is the displacement of the spring.
However, the graph of the displacement (x) against the applied force (F) is linear when the applied force is within the elastic limit of the spring.
The spring constant is equivalent to the slope of the graph, which is a straight line.
Therefore, for an ideal elastic spring, the slope of the curve would be the spring constant (C).
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Galena's specific gravity is 7.5, that of quartz 2.65, and that of liquid mercury 13.6. Given equal-sized samples (volumes) of galena and quartz, which will feel heavier? Choose one: A. galena B. The same volume of water will feel heavier than both of them. C. They will feel about equal. D. quartz
Given equal-sized samples (volumes) of galena and quartz, the Galena sample will feel heavier because of its higher specific gravity. Thus, the correct option is A.
What is the Specific gravity of a substance?Specific gravity is the ratio of the density of a substance to the density of a standard substance in physics. It's typically applied to liquids and solids, but it may also be applied to gases. The most often utilized standard material for liquids and solids is water at 4°C. A substance's specific gravity is dimensionless and is often represented by the Greek symbol ρ.
Relative Density of the given substances:
Galena's specific gravity is 7.5, Quartz's specific gravity is 2.65, and Liquid mercury's specific gravity is 13.6. An object with a specific gravity greater than 1 sinks in water, while one with a specific gravity less than 1 floats in water. The specific gravity of water is 1.0. An object with a specific gravity greater than 1 sinks in water, while one with a specific gravity less than 1 floats in water.
We can conclude from the values above that liquid mercury is heavier than galena, which is in turn heavier than quartz. Therefore, since both quartz and galena are being measured with equal sizes or volumes, galena will feel heavier than quartz.
Therefore, the correct option is A.
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To stretch a spring 5.00cm from its unstretched length, 19.0J of work must be done.1- what is the force constant of the spring ?2- What magnitude force is needed to stretch the spring 5.00cm from its unstretched length?3- How much work must be done to compress this spring 4.00 cm from its unstretched length?4-What force is needed to stretch it this distance?
1) The force constant of the spring is 0.76N/cm, 2) The magnitude force needed to stretch the spring 5.00cm from its unstretched length is 3.80N, 3) Work done to compress this spring 4.00 cm from its unstretched length is 12.48J, 4) Force needed to stretch it this distance is 3.04N.
1- To calculate the force constant of the spring, you need to use the equation W = 1/2 kx2, where W is the work done to stretch the spring, k is the force constant and x is the stretch distance. In this case, W = 19.0J and x = 5.00cm, so k = 19.0/25 = 0.76N/cm.
2- To calculate the magnitude of the force needed to stretch the spring 5.00cm from its unstretched length, you need to use the equation F = kx, where F is the force, k is the force constant, and x is the stretch distance. In this case, F = 0.76N/cm x 5.00cm = 3.80N.
3- To calculate the work done to compress this spring 4.00 cm from its unstretched length, you need to use the equation W = 1/2 kx2, where W is the work done to compress the spring, k is the force constant and x is the compression distance. In this case, W = 1/2 x 0.76N/cm x (4.00 cm)2 = 12.48J.
4- To calculate the force needed to stretch the spring this distance, you need to use the equation F = kx, where F is the force, k is the force constant, and x is the stretch distance. In this case, F = 0.76N/cm x 4.00cm = 3.04N.
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during a one-second period, air is added into a rigid tank. the volume of the tank is 3 m3 and the initial density of air is 1.2 kg/m3; at the end of the charging process, the density of air reaches 6.3 kg/m3. what is the mass flow rate of air that is entering the tank?
The mass flow rate of air that is entering the tank is 15.3 kg/s.
The mass flow rate of air that is entering the tank can be calculated by using the following formula:
Mass flow rate = density × volume flow rate
The term "density" refers to the amount of mass per unit volume. It is calculated as the mass of an object divided by its volume. Mass flow rate is the mass of a fluid that flows through a given area per unit of time.
The volume of the tank is 3 m³.
The initial density of air is 1.2 kg/m³.
At the end of the charging process, the density of air reaches 6.3 kg/m³.
We will first find the volume flow rate.
The volume flow rate is equal to the change in volume over time.
Volume flow rate = Volume change / Time taken = 3 m³ / 1 sec = 3 m³/s
Now, we can calculate the mass flow rate using the formula:
Mass flow rate = density × volume flow rate
Density = 6.3 kg/m³ − 1.2 kg/m³ = 5.1 kg/m³
Mass flow rate = 5.1 kg/m³ × 3 m³/s = 15.3 kg/s
Therefore, the mass flow rate of air entering the tank is 15.3 kg/s.
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A block of mass m is at rest at the origin at t=0. It is pushed with constant force F0 from x=0 to x=Lacross a horizontal surface whose coefficient of kinetic friction is μk=μ0(1−x/L). That is, the coefficient of friction decreases from μ0 at x=0 to zero at x=L.
Part A
We would like to know the velocity of the block when it reaches some position x. Finding this requires an integration. However, acceleration is defined as a derivative with respect to time, which leads to integrals with respect to time, but the force is given as a function of position. To get around this, use the chain rule to find an alternative definition for the acceleration ax that can be written in terms of vx and dvxdx. This is a purely mathematical exercise; it has nothing to do with the forces given in the problem statement.
Express your answer in terms of the variables vx and dvxdx.
I got the answer:
ax =
dvxdxvx
And this was correct, but Im having trouble with Part B:
Now use the result of Part A to find an expression for the block's velocity when it reaches position x=L.
Express your answer in terms of the variables L, F0, m, μ0, and appropriate constants.
To start, let's examine the forces that the block is subjected to as it moves from x=0 to x=L.
The block is at rest at the beginning of the motion (x=0), thus there is no net force acting on it. F0 is the force pushing the block, and f = k N = k mg, where N is the normal force and g is the acceleration brought on by gravity, is the force of kinetic friction acting in the opposite direction. The block is stationary, thus we have:
F0 - μ0 mg = 0
The force pushing the block must thus be equal to and in opposition to the force of friction.
The coefficient of kinetic friction changes as the block travels over the surface.
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