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Vera_Pavlovna [14]
3 years ago
14

Suppose you drop a tennis ball from a height of 6 feet. After the ball hits the floor, it rebounds to 80% of

Physics
2 answers:
statuscvo [17]3 years ago
7 0

Answer: The height after the third bounce is 3.1 meters

Explanation: We know that after every bounce the ball rebounds to 80% (or 0.8 in decimal form) of its previous hight. So if the first height was 6m, after the first bounce, the height will be:

H1 = 6m*0.8 = 4.8m

after the second bounce we have:

H2 = 4.8m*0.8 = 3.84m

and in the third bounce, the height is:

H3 = 3.84m*0.8 = 3.072m

Now we need to round it to the nearest thenth, this is the first number after the decimal point.

We can see that the second number after the decimal point is a 7, so we should round up, and get H3 = 3.1m

Another way to model this situation is with the equation:

H(b) = 6m*0.8^b

Where b is the number of bounces that the ball did.

in this situation, the third bounce is when b= 3, and we have that:

H(3) = 6m*0,8^3 = 3.072m, which is the same result as before.

sasho [114]3 years ago
3 0
The formula to solve this is (.80)^3 X 6 and the answer would be 3.1 feet. That is how high the ball will rebound after its third bounce. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
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3 years ago
A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy
vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J      

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

a)

We know that kinetic energy KE is given as follows

KE=\dfrac{1}{2}mu^2

m=mass

u=velocity

Now by putting the values in the above equation we get

KE=\dfrac{1}{2}\times 65\times 5.2^2\ J

KE=878.8 J

b)

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2

Now by putting the values in the above equation we get

W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0
3 years ago
A motor running at 2600 rev/min is suddenly switched off and decelerates uniformly to rest after 10 s. Find the angular decelera
Mama L [17]
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change in angular speed = (zero - 2,600 RPM)  =  -2,600 RPM

time for the change = 10 sec

Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec

(-260 rev/min-sec) x (1 min / 60 sec) = <em>-(4 1/3)  rev / sec²</em>

Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .

The average speed is  1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.

Number of revs = (average speed) x (time) = (21 2/3) x (10sec) = <em>(216 2/3) revs</em>
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3 years ago
What affect does friction have on motion
satela [25.4K]

Answer: friction reduces the speed during motion

Explanation:

The more the friction, the lesser the speed during motion

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3 years ago
1 point
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Answer:

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