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rusak2 [61]
4 years ago
13

A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo

unced to a height of 8m, how would youdescribe the collision between the ball and the floor? b. If the ball had not bounced at all, how would you describethe collision between the ball and the floor? c. What happened to the energy lost by the ball during thecollision?
Physics
1 answer:
kifflom [539]4 years ago
5 0

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

You might be interested in
Two forces,
serg [7]

First compute the resultant force F:

\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N

\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N

\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N

Then use Newton's second law to determine the acceleration vector \mathbf a for the particle:

\mathbf F=m\mathbf a

(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a

\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}

Let \mathbf x(t) and \mathbf v(t) denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so \mathbf v(0)=0. Then the particle's velocity vector at <em>t</em> = 10.4 s is

\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du

\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}

\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}

If you don't know calculus, then just use the formula,

v_f=v_i+at

So, for instance, the velocity vector at <em>t</em> = 10.4 s has <em>x</em>-component

v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}

(b) Compute the angle \theta for \mathbf v(10.4\,\mathrm s):

\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ

so that the particle is moving at an angle of about 313º counterclockwise from the positive <em>x</em> axis.

(c) We can find the velocity at any time <em>t</em> by generalizing the integral in part (a):

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

Then using the fundamental theorem of calculus again, we have

\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du

where \mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m is the particle's initial position. So we get

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du

\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}

\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m

So over the first 10.4 s, the particle is displaced by the vector

\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0
3 years ago
A piece of Nichrome wire has a radius of 7.1 x 10-4 m. It is used in a laboratory to make a heater that dissipates 3.20×102 W of
bekas [8.4K]

Answer:

the length of the Nichrome wire is <em>59.88 m</em>

<em></em>

Explanation:

The resistance of a wire is determined by:

P = V² / R                              or    R = V² / P

where

  • V is the voltage
  • R is the resistance
  • P is the power

We also know that

R = PL / A

where

  • A is the area and it is equal to πr²
  • P = 3.20 × 10² W

Therefore, the length of the Nichrome wire is

L  = RA / ρ

L = (V²/P)(πr²) / ρ

L = V²πr² /  ρ P

L = (110 V)²(π)(7.1 × 10⁻⁴ m)² / (3.2 × 10² W)(100 × 10⁻⁸ Ω·m)

<em>L = 59.88 m</em>

7 0
4 years ago
Two train whistles, and , each have a frequency of 392 Hz. is stationary and is moving toward the right (away from ) at a speed
trasher [3.6K]

Answer:

Part a)

f = 371.1 Hz

Part b)

f = 417.7 Hz

Part c)

beat frequency = 46.6 Hz

Explanation:

Part a)

Due to doppler's Effect the frequency of the sound heard by the train which is moving away from the observer is given as

f_1 = f_0\frac{v + v_o}{v + v_s}

f_1 = 392(\frac{340 + 15}{340 + 35})

f_1 = 371.1 Hz

Part b)

Now from the second train which is approaching the person we can say

f_2 = f_0\frac{v - v_o}{v - v_s}

f_2 = 392(\frac{340 - 15}{340 - 35})

f_2 = 417.7 Hz

Part c)

As we know that beat frequency is the difference in the frequency from two sources

f_b = f_2 - f_1

f_b = 417.7 - 371.1 = 46.6 Hz

8 0
3 years ago
A 100 kilogram person acquires a velocity of 15 meters per second down a ski slope.What is the skier's kinetic energy?
azamat

11250J

Explanation:

Given parameters:

Mass of the person = 100kg

Velocity = 15m/s

Unknown:

Kinetic energy  = ?

Solution:

The kinetic energy of a body is energy due to the motion of the body.

 Kinetic energy = \frac{1}{2} m v²

    m is the mass of the body

    v is the velocity

  Substituting the parameters:

     

 K. E =  \frac{1}{2} x 100 x 15² = 11250J

Learn more:

Kinetic energy brainly.com/question/6536722

#learnwithBrainly

4 0
3 years ago
John(body mass=160pounds) is taking off for a long jump. The average ground reaction force Fg at takeoff is 1400 N pointing forw
slega [8]

Answer:

The free body diagram of John is shown in the attached figure (in the FBD john's mass is supposed to be concentrated at his center of mass and FBD is made of center of mass)

b) As shown in the FBD the ground reaction forces are:

i) In X direction F_{x}=1400cos(35^{o})=1146.81N

ii) In Y direction F_{y}=1400sin(35^{o})=803.0N

c) The respective accelerations in x and y direction's is calculated by newton's second law as indicated under

\sum F_{x}=ma_{x}\\\\\therefore a_{x}=\frac{\sum F_{x}}{m}=\frac{1146.8N}{72.57kg}=15.80m/s^{2}\\\\\sum F_{y}=ma_{y}\\\\\therefore a_{y}=\frac{\sum F_{y}}{m}=\frac{803.00-72.57\times 9.81}{72.57}=1.255m/s^{2}

4 0
3 years ago
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