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scZoUnD [109]
3 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:
Crank3 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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zzz [600]

Answer:

compared to the incident angle, the refracted angle is 45.56⁰

Explanation:

From Snell's law;

n₁sin(I) = n₂sin(r)

Where;

n₁ is the refractive index of light in medium 1 = 1.2

n₂ is the refractive index of light in medium 2 = 1.4

I is the incident angle

r is the refractive angle

n = \frac{1}{sin(I)}\\\\sin(I) = \frac{1}{n}\\\\sin(I) =\frac{1}{1.2}\\\\sin(I) =0.8333\\\\I = sin^-{(0.8333)

I = 56.439⁰

Applying snell's law

n_1sin(I) = n_2sin(r)\\\\sin(r) = \frac{n_1sin(I) }{n_2}\\\\sin(r) = \frac{1.2*sin(56.439) }{1.4}\\\\sin(r) = 0.714\\\\r = sin^-(0.714)\\\\r = 45.56^o

Therefore, compared to the incident angle, the refracted angle is 45.56⁰

3 0
3 years ago
What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m
andre [41]

Answer:

Velocity of a proton, v=1.7\times 10^6\ m/s    

Explanation:

It is given that,

Potential difference, V=15\ kV=15\times 10^3\ V

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

\dfrac{1}{2}mv^2=qV

q is the charge of proton

m is the mass of proton

v=\sqrt{\dfrac{2qV}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}

v=1695361.75\ m/s

v=1.69\times 10^6\ m/s

or

v=1.7\times 10^6\ m/s

So, the velocity of a proton is 1.7\times 10^6\ m/s. Hence, this is the required solution.

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What was done to test if Comets are responsible for 1/2 the Earth's water?
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✯Hello✯

↪  A satellite was crashed into a comet (on purpose of course)

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Since the bag was at rest, its initial momentum is zero. The velocity of the ball before collision is 500 ms-1.

<h3>Linear momentum</h3>

The term momentum in physics refers the product of mass and velocity. If we know mass of the object and its velocity, then we calculate the momentum.

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The velocity of the bullet before collision =  0.01 kg × v + 0 =  5 Kgms-1

v = 5 Kgms-1/0.01 kg

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Learn more about momentum: brainly.com/question/904448

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