Answer:
Here its right but its also better than Barney's response
Explanation:
W, Y, Z, X or C
Answer:
W, Y, Z, X
Explanation:
When the bowling ball has fallen halfway down the building (height = 20 m), it has a speed of 19.8 m/s.
How much potential energy does the bowling ball have?
How much kinetic energy does the bowling ball have?
How much total energy (potential + kinetic) does the bowling ball have?
Of the bowling ball’s total energy, is more in the form of potential or kinetic energy?
Answer:
I think the answer is 19.8 potential energy
Explanation:
NONE.
take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.
pls give me ideas of what to take a photo of for this I'm really stuck :(
please help asap!!
Explain the movement of a roller coaster in terms of potential and kinetic energy? When are these energies thegreatest? Smallest? Are they ever the same?
Answer:
Potential energy: Greatest at the top of the hill
Kinetic energy: Greatest at the bottom of the hill
The two meet at some point on the way down!
Explanation:
Potential energy is energy that represents an object's potential for motion. Kinetic energy is that object's energy during motion. They're two sides of the same coin, and in fact, their sum gets a special name: mechanical energy. Potential energy builds up in reaction to working against certain forces - in the case of the roller coaster, that primary force is gravity. Gravity exerts a downward force on the roller coaster, and it takes work to pull it up the hill.
When it reaches the peak, the coasters potential energy is at its highest, and the moment it crests over the hill and begins its descent, that gravitational potential energy starts converting into kinetic energy: the coaster starts accelarating down the track, and the potential energy decreases at the same rate that the kinetic energy increases.
At the bottom of the hill, all of that potential energy has become kinetic energy, and the coaster zooms along the track, hopefully not giving too many riders nausea
What kind of scattering (Rayleigh, Mie, or non-selective) would you expect to be most important when radiation of the specified wavelength encounters the following natural or anthropogenic particles?
Slides 16-31, Lecture 2 ought to help - slides 19, 24, and 31 are key.
Wavelength O2 molecules Smoke particles Cloud droplets Rain droplets
(size 10^-10 m) (size 0.3 (μm) (20 μm) (size 3 mm)
550 nm
11 μm
1600 nm
1 cm
Solution :
1. Rayleigh scattering takes place when the particle size is smaller than the wavelength (λ).
2. Mie scattering takes place when particle size is nearly equal to the wavelength (λ).
3. Non-selective scatter takes place when particle size in greater than the wavelength (λ).
We have the sizes of different particles :
[tex]$O_2 \rightarrow 10^{10} \ m $[/tex]
Smoke particles [tex]$\rightarrow 3 \times 10^{-7} \ m$[/tex]
Cloud droplets [tex]$\rightarrow 2 \times 10^{-5} \ m$[/tex]
Rain droplets [tex]$\rightarrow 3 \times 10^{-3} \ m$[/tex]
Wavelength [tex]$ O_2 $[/tex] Smoke particles Cloud droplets Rain droplets
[tex]$10^{-10} \ m$[/tex] [tex]$ 3 \times 10^{-7} \ m$[/tex] [tex]$ 2 \times 10^{-5} \ m$[/tex] [tex]$ 3 \times 10^{-3} \ m$[/tex]
[tex]$5500 \times 10^{-4} \ m$[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$11 \times 10^{-6} \ m $[/tex] Rayleigh Rayleigh Non-selective Non-selective
[tex]$1600 \times 10^{-10} \ m $[/tex] Rayleigh Non-selective Non-selective Non-selective
[tex]$10^{-2} \ m $[/tex] Rayleigh Rayleigh Rayleigh Mie
A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it hits the wall, it is moving at 3.1 m/s. How high is the wrecking ball when it hits the wall? (Show your work and follow all of the steps of the GUESS method. Check your answer after you submit the form - it's in the feedback for this question.) |
Einstein's equivalence principle says that __________. Einstein's equivalence principle says that __________. everyone measures the speed of light to be equivalent someone traveling at 0.9c will age at the same rate as someone at 0.99c all people see themselves at an equivalent distance to the center of the universe the effects of gravity are exactly equivalent to the effects of acceleration
Answer:
Einstein's equivalence principle says that __________.
the effects of gravity are exactly equivalent to the effects of acceleration
Explanation:
The equivalence principle is one of the fundamental laws of physics, as enunciated by Einstein. It categorically states that the gravitational and inertial forces are of a similar nature. In physics, a gravitational acceleration is the acceleration of an object in a free fall within a space. The importance of Einstein's Equivalence Principle is explained by his theory of general relativity. This theory states that mass is the same, whether inertial or gravitational.
According to the Einstein's equivalence principle, the effects of gravity are exactly equivalent to the effects of acceleration.
Einstein's equivalence principle says that the effects of gravity are exactly equivalent to the effects of acceleration.
What is Einstein's equivalence principle?Einstein's equivalence principle states that the the force due to gravity and the force of inertia are similar in the nature and there is no need to distinct them.
The inertia force is opposite in direction to accelerating force of a body. Thus the Einstein's equivalence principle can also be stated as "the effects of gravity are exactly equivalent to the effects of acceleration." Form the given option the correct option which can be filled in the blank is option 2 which states that the effects of gravity are exactly equivalent to the effects of acceleration.Thus Einstein's equivalence principle says that the effects of gravity are exactly equivalent to the effects of acceleration.
Learn more about the Einstein's equivalence principle here
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1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) the time taken for it to return to 0 (c) the taken for it to be 35m below 0
Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is
v = 30 m/s - g t
where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by
y = (30 m/s) t - 1/2 g t ²
(a) At its maximum height, the particle has 0 velocity, which occurs for
0 = 30 m/s - g t
t = (30 m/s) / g ≈ 3.06 s
at which point the particle's maximum height would be
y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m
(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.
(c) The particle falls 35 m below its starting point when
-35 m = (30 m/s) t - 1/2 g t ²
Solve for t to get a time of about t ≈ 7.1 s
Car À moves at a speed of 8m/s for 43 seconds. Car B moves at a speed of 7 m/s for 50 seconds. Which car traveled a longer distance
Please show working
Distance = (speed) x (time)
Car A: Distance = (8 m/s) x (43 s) = 344 meters
Car B: Distance = (7 m/s) x (50 s) = 350 meters
350 meters is a longer distance than 344 meters.
Car-B traveled a longer distance than Car-A did.
Answer:
[tex]\boxed {\boxed {\sf Car \ B : 350 \ meters }}[/tex]
Explanation:
Distance is equal to the product of speed and time.
[tex]d=s*t[/tex]
1. Car A
Car A has a speed of 8 meters per second and travels for 43 seconds.
[tex]s= 8 \ m/s \\t= 43 \ s[/tex]
Substitute the values into the formula.
[tex]d= 8 \ m/s *43 \ s[/tex]
Multiply and note that the seconds will cancel out.
[tex]d= 8 \ m*43= 344 \ m[/tex]
2. Car B
Car B has a speed of 7 meters per second and travels for 50 seconds.
[tex]s= 7 \ m/s \\t= 50 \ s[/tex]
Substitute the values in and multiply.
[tex]d= 7 \ m/s * 50 \ s[/tex]
[tex]d= 7 \ m * 50 = 350 \ m[/tex]
350 meters is a longer distance than 344 meters, so Car B traveled the longer distance.
An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision
Answer:
9.3m/s
Explanation:
Based on the law of conservation of momentum
Sum of momentum before collision = sum of momentum after collision
m1u1 +m2u2 = m1v1+m2v2
m1 = 8kg
u1 = 15.4m/s
m2 = 10kg
u2 = 0m/s(at rest)
v1 = 3.9m/s
Required
v2.
Substitute
8(15.4)+10(0) = 8(3.9)+10v2
123.2=31.2+10v2
123.2-31.2 = 10v2
92 = 10v2
v2 = 92/10
v2 = 9.2m/s
Hence the velocity of the 10.0 kg object after the collision is 9.2m/s
Name and explain the
various types of friction.
Answer:
There are four types of friction: static, sliding, rolling, and fluid friction. Static, sliding, and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction, and then rolling friction, which is weakest. Fluid friction occurs in fluids, which are liquids or gases.
Explanation:
As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock. As the waves recede, they carry the sediment away. In this scenario, which process represents weathering, and which process represents erosion?
Answer:
WEATHERING is represented by the scenario (As waves crash into rock along the shoreline, particles of sand, shell, and other materials in the ocean water loosen tiny bits of sediment from the rock).
Erosion is represented by the scenario (As the waves recede, they carry the sediment away).
Explanation:
A wave is a disturbance which travels through a medium and transfers energy from one point to another. When wind blows over a water body like the ocean, ocean waves are formed. As the generated energy from the wind is transported through the water by the waves, the can hit against rocks on the shores leading to its break down with time. WEATHERING occurs when tiny bit of sediments from rocks are loosened due to the impact of ocean waves.
Erosion can be described as the wearing away of the earth's surface due to the impact of wind, rainfall ( water) or waves. There are different types of erosion which is classified according it's cause of formation.
Wave erosion occurs when sediments such as sand, shell and other materials are carried to the shoreline by ocean waves. This erodes the shore over time as the sediments act like sandpapers.
One disadvantage to experimental research is that experimental conditions do not always reflect reality.
Please select the best answer from the choices provided
T
F
Answer:
It's true I took the test on Edge.
Explanation:
Answer:
True
Explanation:
Got it right on edg
"45 meters north" is an example of
Answer:
Displacement
Explanation:
The quantity 45m north is a typical example of displacement.
Displacement is the distance traveled by a body in a specific direction. Displacement is a vector quantity with both magnitude and direction.
When we are specifying the displacement of a body, the direction must be indicated accurately. Therefore, the quantity given is displacementWhat Coulombs discovered almost 300
years ago
Answer:
ummm hehe this is my time to shine
Explanation:
MERICIA!!!!!!!!!!!!!!!!!!!!!!!
What is Ex(P), the value of the x-component of the electric field produced by by the line of charge at point P which is located at (x,y) = (a,0), where a = 8.7 cm?
Answer:
The answer is below
Explanation:
We are going to use Gauss’ law to find the electric field equation. Since electric field is coming from an infinite line of charge, hence it is going out in a radial direction.
Therefore we use the area of the electric field which passes through, forming a Gaussian cylinder. We neglect the ends of the area.
Hence:
[tex]\int\limits {E} \, dA=\frac{Q_{enc}}{\epsilon_o}\\\\E(2\pi rL)= \frac{\lambda L}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r\epsilon_o} \\\\Given \ that:\\\\r=a=8.7\ cm=0.087\ m, \lambda=-2.3 \mu C/cm=-2.3*10^{-4}\ C/m,\epsilon_o=8.85*10^{-12}F/m.\\\\Hence:\\\\E=\frac{-2.3*10^{-4}}{2\pi *0.087*8.85*10^{-12}}=-4.75*10^7\ N/C[/tex]
The value of the x-component of the electric field is -475213.968 newtons per coulomb.
Procedure - Determination of the magnitude of an electric field at a given pointIn this question we shall apply Gauss' Law to determine the magnitude of the electric field ([tex]E_{x}[/tex]), in newtons per coulomb, rapidly and based on the assumptions of uniform charge distribution and cylindrical symmetry.
[tex]\frac{Q_{enc}}{\epsilon_{o}} = \oint\,\vec E\,\bullet d\vec A[/tex] (1)
Where:
[tex]Q_{enc}[/tex] - Enclosed charge, in coulombs.[tex]\epsilon_{o}[/tex] - Vacuum permitivity, in quartic second-square amperes per kilogram-cubic meter.[tex]\vec E[/tex] - Electric field vector, in newtons per coulomb.[tex]\vec A[/tex] - Area vector, in square meters.Based on all assumptions, we simplify (1) as follows:
[tex]\frac{\lambda\cdot l}{\epsilon_{o}} = E \cdot (2\pi\cdot r\cdot l)[/tex]
And the equation of the x-component of the electric field is:
[tex]E = \frac{\lambda}{2\pi\cdot \epsilon_{o}\cdot r}[/tex] (2)
Where [tex]\lambda[/tex] is the linear charge density, in coulomb per meter.
If we know that [tex]\lambda = -2.3\times 10^{-6}\,\frac{C}{m}[/tex] and [tex]a = 0.087\,m[/tex], then the electric field produced by the line of charge at point P is:
[tex]E = \frac{\left(-2.3\times 10^{-6}\,\frac{C}{m} \right)}{2\pi\cdot \left(8.854\times 10^{-12}\,\frac{s^{4}\cdot A^{2}}{kg\cdot m^{3}} \right)\cdot (0.087\,m)}[/tex]
[tex]E_{x} = -475213.968 \,\frac{N}{C}[/tex]
The value of the x-component of the electric field is -475213.968 newtons per coulomb. [tex]\blacksquare[/tex]
RemarkThe figure is missing, we present the corresponding image in the file attached below.
To learn more on electric fields, we kindly invite to check this verified question: https://brainly.com/question/12757739
If there is "waste" energy, does the Law of Conservation of Energy still apply?
Explanation:
Yes, the law of conservation of energy still applies even if there is waste energy.
The waste energy are the transformation products of energy from one form to another.
According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".
But of then times, energy is lost as heat or sound within a system.
If we take into account these waste energy, we can see that energy is indeed conserved. The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.Two metal bricks are held off the edge of a balcony from the same height above the ground. The bricks are the same size but one is made of Titanium (density of 4.5 g/cm%) and one is made of Lead (density of 11.3 g/cm3) so the Lead is about twice as heavy as the Titanium. The time it takes the bricks to reach the ground will be:________.
a. less but not necessarily half as long for the heavier brick
b. about half as long for the lighter brick
c. less but not necessarily half as long for the lighter brick
d. about half as long for the heavier brick
e. about the same time for both bricks
Answer:
e.
Explanation:
Assuming that the air resistance is neglectable, both bricks are only accelerated by gravity, which produces a constant acceleration on both bricks, which is the same, according Newton's 2nd Law, as we can see below:[tex]F_{g} = m*g = m*a (1)[/tex]⇒a = g = 9.8m/s² (pointing downward)Since acceleration is constant, if both fall from the same height, we can apply the following kinematic equation:[tex]\Delta y = v_{o} * t - \frac{1}{2} *g*t^{2} (2)[/tex]
Since both bricks are held off the edge, the initial speed is zero, so (2) reduces to the following equation:[tex]h =\frac{1}{2} *g*t^{2} (3)[/tex]
Since h (the height of the balcony) is the same, we conclude that both bricks hit ground at exactly the same time.If the air resistance is not negligible, due both bricks have zero initial speed, and have the same shape, they will be affected by the drag force in similar way, so they will reach the ground at approximately the same time.what is the direction of the third force that would cause the box to remain stationary on the ramp ?
An arrow pointing from the bottom of the ramp to the top, I assume it would be friction.
which experimental result led to a revision of Thomas's plum pudding model of the atom?
A. electrons were found to have higher energy the farther they are from the nucleus
B. the beam in a cathode ray tube was moved by an electric force
C. A few alpha particles bounced off a thin sheet of gold foil
D. most alpha particles passed straight through a thin sheet of gold foil
Answer: C. A few alpha particles bounced off a thin sheet of gold foil.
A truck travelling down the street suddenly brakes, applying a 14 N force over 3.5 seconds. What was the impulse over the given time.
Answer:
49 Ns
Explanation:
Given data
Force= 14N
time = 3.5seconds
Applying the expression for impulse
P= Ft
substitute
P=14*3.5
P=49 Ns
Hence the impulse is 49 Ns
A vertical piston-cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter for 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will doublethe pressure of the gas inside the cylinder.
Answer:
the local atmospheric pressure is 93.63 kPa
the mass of the weights is 156.9 kg
Explanation:
Given that;
Initial pressure of gas = 100 kPa
mass of piston = 10 kg and diameter = 14 cm = 0.14 m
g = 9.81 m/s²
Now,
P_gas = P_atm + P_piston
100 = P_atm + P_piston --------- let this equation 1
P_piston = M_piston × g / A = (10 × 9.81) / π/4×d²
P_piston = 98.1 / (π/4×( 0.14 )²)
P_piston = 98.1 / 0.01539 = 6374,269 Pa = 6.37 kPa
now, from equation 1
100 = P_atm + P_piston
we substitute
100 = P_atm + 6.37
P_atm = 100 - 6.37
P_atm = 93.63 kPa
Therefore, the local atmospheric pressure is 93.63 kPa
Now for pressure of the gas in the cylinder ⇒ 2×initial pressure
Pgas_2 = 2 × 100 = 200 kPa
Pgas_2 = P_atm + P_piston + P_weight
Pgas_2 = P_gas + P_weight
we substitute
200 kPa = 100 kPa + P_weight
P_weight = 200 kPa - 100 kPa
P_weight = 100 kPa = 100,000 Pa
Also;
P_weight = M×g / A
100,000 Pa = ( M × 9.81 ) / (π/4 × (0.14)²)
100,000 × 0.01539 = M × 9.81
1539 = M × 9.81
M = 1539 / 9.81
M = 156.9 kg
Therefore, the mass of the weights is 156.9 kg
why do players choose to follow the unconventional route of kicking down the middle
Answer:
My biggest reason is to make it a habit. Even if the ball goes into the endzone it is a live ball and the offensive players must down the ball. Don't leave any room for "I thought he downed it" or "I thought I heard the whistle" just run to the ball always.
If the players slow down and the returner takes it out of the end zone it could be a big return. Players are on a full sprint for 40+ yards sometimes and instead of breaking down, they choose to contine through the goal line to slow down at a decreased rate (possibly limiting a muscle pull injury).
Which of the following is a vector quantity?
speed
distance
acceleration
Two steamrollers begin 105 mm apart and head toward each other, each at a constant speed of 1.20 m/s. At the same instant, a fly that travels at a constant speed of 2.50 m/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision.
Required:
What distance does the fly travel?
Answer: 109.4 mm
Explanation: Distance is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time: d = vt
The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:
location of collision = [tex]\frac{0.105}{2}[/tex] = 0.0525 m
To reach that point, both steamrollers will have spent
[tex]v=\frac{\Delta x}{t}[/tex]
[tex]t=\frac{\Delta x}{v}[/tex]
[tex]t=\frac{0.0525}{1.2}[/tex]
t = 0.04375 s
The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:
d = 2.5*0.04375
d = 0.109375 m
Until it is crushed, the fly will have traveled 109.4 mm.
A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation
Answer:
Explanation:
The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .
1/2 k x² = mgx
.5 x k x .33² = m x 9.8 x .33
k / m = 59.4
frequency of oscillation = [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]
= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]
= 1.22 per second .
A person pushes down on a lever with a force of 100 N. At the other end of the lever, a force of 200 N lifts a heavy object. What is the mechanical advantage of the lever?
A. 1/2, because the object will be lifted half the distance
B. -1, because the direction changes
C. 2, because the output force is twice the input force
D. 1, because the same amount of work is done
Answer:
Explanation:
C 200÷100=2
Output ÷ Input= MA
Which of the physical variables listed below will change when you change the area of the capacitor plates (while keeping the battery connected).
a. Capacitance
b. Charge on the plates
c. Voltage across the plates
d. Net electric field between the plates
e. Energy stored in the capacitor
Answer:
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
Explanation:
Let A be the area of the capacitor plate
The capacitance of a capacitor is given as;
[tex]C = \frac{Q}{V} = \frac{\epsilon _0 A}{d} \\\\[/tex]
where;
V is the potential difference between the plates
The charge on the plates is given as;
[tex]Q = \frac{V\epsilon _0 A}{d}[/tex]
The energy stored in the capacitor is given as;
[tex]E = \frac{1}{2} CV^2\\\\E = \frac{1}{2} (\frac{\epsilon _0 A}{d} )V^2[/tex]
Thus, the physical variables listed that will change include;
a. Capacitance
b. Charge on the plates
e. Energy stored in the capacitor
Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.
a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?
Answer:
a) ω₁ = ω₂ = 3.7 rad/sec
b) Δθ₁ = Δθ₂ = 18.5 rad
c) d₁ = 14.5 m d₂ = 57.5 m
d) Fc1 = 273.9 N Fc2 = 1069.8 N
e) The boy near the outer edge.
Explanation:
a)
Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.However, linear speeds of points at different distances from the center, are different.Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:[tex]v = \omega*r (1)[/tex]
Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:[tex]\omega = \frac{v_{out} }{r_{out} } = \frac{11.5m/s}{3.14m} = 3.7 rad/sec (2)[/tex]
As we have already said, ωout = ωin = 3.7 rad/secb)
Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:[tex]\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)[/tex]⇒ Δθ₁ = Δθ₂ = 18.5 rad.
c)
The linear distance traveled by each child, will be related with the linear speed of them.Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:[tex]v_{inn} = \omega * r_{inn} = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)[/tex]
vout is a given of the problem ⇒ vout = 11. 5 m/s
Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:[tex]d_{inn} = v_{inn} * t = 2.9m/s* 5.0 s = 14.5 m (5)[/tex]
[tex]d_{out} = v_{out} * t = 11.5 m/s* 5.0 s = 57.5 m (6)[/tex]
d)
The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:[tex]F_{c} = m*\frac{v^{2}}{r} (7)[/tex]
Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:[tex]F_{cin} = m*\frac{v_{in}^{2}}{r_{in}} = 25.4 kg* \frac{(2.9m/s)^{2} }{0.78m} = 273.9 N (8)[/tex]
In the same way, we get Fcout (the force on the boy near the outer edge):[tex]F_{cout} = m*\frac{v_{out}^{2}}{r_{out}} = 25.4 kg* \frac{(11.5m/s)^{2} }{3.14m} = 1069.8 N (9)[/tex]
e)
The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.The maximum friction force is given by the product of the coefficient of static friction times the normal force.Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.As both boys have the same mass, the normal force is also equal.This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:[tex]F_{frs} = \mu_{s} * m* g (10)[/tex]If this force is greater than the centripetal force, the boy will be able to hold on.So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.In which number are the zeros not significant?
100.0
O 0.0003
O 4.00005
O 1.0004
Answer:
0.0003
Explanation:
In the rules of Sig Figs, all zeros before with decimals are not sigificant. I.E. 0.00000000000000009. Despite how many 0's there are, only the 9 is significant. Zeros before a number is not significant. In 100, only the one is signficant in 100. with a dot at the end, the one and the two zeros are significant. hope this helps.
Answers:
the second option
Explanation:
What happens to kinetic energy when you decrease the velocity of a moving object?
