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scZoUnD [109]
4 years ago
15

A super ball is dropped from a height of 100 feet. Each time it bounces, it rebounds half the distance it falls. How many feet w

ill the ball have traveled when it hits the ground for the fourth time
Physics
1 answer:
Crank4 years ago
4 0

The total distance travelled by the ball after the fourth impact is 275 feet.

<u>Explanation:</u>

Given-

Height, h = 100 feet

Rebounds half the distance

Distance in feet for the fourth time, x = ?

For the first time, the distance travelled by the ball is, x = 100 feet

For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)

So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet

For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)

So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet

For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)

So, the distance travelled after the fourth impact, x = 250 + 12.5 + 12.5 = 275 feet

Therefore, total distance travelled by the ball after the fourth impact is 275 feet.

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A uniform disk with radius 0.650 m
VashaNatasha [74]

Answer:

a = 13.758\,\frac{m}{s^{2}}

Explanation:

First, the instant associated to the angular displacement is:

(1.10\,\frac{rad}{s} )\cdot t + (6.30\,\frac{rad}{s^{3}} )\cdot t^{2} - 0.628\,rad = 0

Roots of the second-order polynomial are:

t_{1} \approx 0.240\,s, t_{2} \approx -0.415\,s

Only the first root is physically reasonable.

The angular velocity is obtained by deriving the angular displacement function:

\omega (0.240\,s) = 1.10\,\frac{rad}{s}+ (12.6\,\frac{rad}{s^{2}})\cdot (0.240\,s)

\omega (0.240\,s) = 4.124\,\frac{rad}{s}

The angular acceleration is obtained by deriving the previous function:

\alpha (0.240\,s) = 12.6\,\frac{rad}{s^{2}}

The resultant linear acceleration on the rim of the disk is:

a_{t} = (0.650\,m)\cdot (12.6\,\frac{rad}{s^{2}} )

a_{t} = 8.190\,\frac{m}{s^{2}}

a_{n} = (0.650\,m)\cdot (4.124\,\frac{rad}{s} )^{2}

a_{n} = 11.055\,\frac{m}{s^{2}}

a = \sqrt{a_{t}^{2}+a_{n}^{2}}

a = 13.758\,\frac{m}{s^{2}}

3 0
3 years ago
? Why might the expejiment need to be repeated when the hypothesis
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Detailed research on comets indicates that they cannot be older than HOW MANY years.
Free_Kalibri [48]
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3 years ago
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A scientist spends most of his time observing the night time sky and tracking the orbits of planets. Which branch of physical sc
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4 0
4 years ago
A car is brought to rest in a distance of 484m using a constant acceleration of -8.0m/s^2. What was the velocity of the car when
Agata [3.3K]

Answer:

88 m/s

Explanation:

To solve the problem, we can use the following SUVAT equation:

v^2-u^2=2ad

where

v is the final velocity

u is the initial velocity

a is the acceleration

d is the distance covered

For the car in this problem, we have

d = 484 m is the stopping distance

v = 0 is the final velocity

a=-8.0 m/s^2 is the acceleration

Solving for u, we find the initial velocity:

u=\sqrt{v^2-2ad}=\sqrt{-2(8.0)(484)}=88 m/s

6 0
4 years ago
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