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lana66690 [7]
4 years ago
14

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost

? b) What is the ball's speed after the bounce? c) Where did the energy go?
Physics
1 answer:
tangare [24]4 years ago
4 0
The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
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50 points!!! Kinetics
iris [78.8K]

Answer:

98 m √

Explanation:

How about s = Vo * t + ½at² ?

s = h = Vo * 2s - 4.9m/s² * (2s)² = 2Vo - 19.6

and

h = Vo * 10s - 4.9m/s² * (10s)² = 10Vo - 490

Subtract 2nd from first:

0 = -8Vo + 470.4

Vo = 58.8 m/s

h = 58.8m/s * 2s - 4.9m/s² * (2s)² = 98 m

3 0
3 years ago
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A pendulum is used in a large clock. The pendulum has a mass of 2 kg. If the pendulum is moving at a speed of 2.9 m/s when it re
Rufina [12.5K]
This is a classic example of conservation of energy. Assuming that there are no losses due to friction with air we'll proceed by saying that the total energy mus be conserved.
E_m=E_k+E_p
Now having information on the speed at the lowest point we can say that the energy of the system at this point is purely kinetic:
E_m=Ek=\frac{1}{2}mv^2
Where m is the mass of the pendulum. Because of conservation of energy, the total energy at maximum height won't change, but at this point the energy will be purely potential energy instead.
E_m=E_p
This is the part where we exploit the Energy's conservation, I'm really insisting on this fact right here but it's very very important, The totam energy Em was
E_M=\frac{1}{2}mv^2
It hasn't changed! So inserting this into the equation relating the total energy at the highest point we'll have:
E_p=mgh=E_m=\frac{1}{2}mv^2
Solving for h gives us:
h=\frac{v^2}{2g}.
It doesn't depend on mass!

4 0
3 years ago
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A 40kg rock falls off a cliff that is 50 meters high. How fast is the speed of the rock when it hits the ground
Oksi-84 [34.3K]

31.3m/s

Explanation:

Given parameters:

Mass of rock = 40kg

Height of cliff = 50m

Unknown:

Speed of rock when it hits ground = ?

Solution:

We are going to use the appropriate motion equation to solve this problem

The rock is falling with the aid of gravitational force. The force is causing it to accelerate with an amount of velocity.

  Using;

                      V²  = U² + 2gH

V = unknown velocity

U = initial velocity = O

g = acceleration due to gravity = 9.8m/s²

H = height of fall

since the initial velocity of the bodyg is 0

     V²  = 2gH

    V= √2gH = √2 x 9.8 x 50 = 31.3m/s

learn more:

Velocity brainly.com/question/4460262

#learnwithBrainly

7 0
4 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
A plucked guitar string produces a sound wave with a wavelength of 0.15 m and a velocity of 10.5 m/s, what is the frequency of t
Leno4ka [110]
We know,
V= f× wavelength
10.5= f×0.15
f=10.5/0.15
f= 70 Hz
8 0
3 years ago
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