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tensa zangetsu [6.8K]
3 years ago
8

Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces

off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber.
1. In which case is the change in momentum of the ball between the instant just before the ball collides with the floor or rubber and the instant just after the ball leaves the floor or rubber the biggest?

a. Case 1
b. Case 2
c. Same in both

2. In which case is the average force acting on the ball during the collision the biggest?

a. Case 1
b. Case 2
c. Same in both
Physics
1 answer:
Kitty [74]3 years ago
7 0

Answer:

<em>1. c. Same in both</em>

<em>2. a. Case 1</em>

<em></em>

Explanation:

1. The balls are identical in all sense, which means that if they are dropped from the same height, they should posses the same kinetic energy just before they collide with either the concrete floor or the stretchy rubber. Also, since they reach the same height when they bounced of the concrete floor or the piece of stretchy rubber, it means that they posses the same amount of kinetic energy at this point. Since their kinetic energy at these two points are the same, and they have the same masses, then this means that their momenta at these two instances will also be equal. Since all these is true, then the change in the momentum of the balls between the instance just before hitting the concrete floor or the stretchy rubber material and the instant the ball just leave the floor or the stretchy material is the same for both.

2. The ball that falls on the concrete will experience the greatest force, since the time of impact is small, when compared to the time spent by the other ball in contact with the stretchy rubber material; which will stretch, thereby extending the time spent in contact between them.

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A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi
USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

f=\frac{nv}{2l}\\.

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\.

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz

Hence the require beat is

B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz.

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204

Now to determine the extension,

L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\

4 0
3 years ago
State a Newtowns second law​
astraxan [27]

Answer:

the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a. For an external applied force, the change in velocity depends on the mass of the object.

5 0
3 years ago
an object is moving with initial velocity of 5 m/s. After 10 seconds final velocity is 10 m/s. Calculate its acceleration.​
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Answer:

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Explanation:

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What should be done to lift the same load by applying less effort on an inclined plane​
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Answer:

Reduce the friction at the surface

Explanation:

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2 years ago
a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what fo
12345 [234]

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

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Solution:

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Where:

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Substitute W = 80.2\ J\ and\ d =25.0\ m in work done formula.

80.2 = F\times 25

F=\frac{80.2}{25}

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

3 0
3 years ago
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