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Sauron [17]
3 years ago
10

Children slide down a frictionless water slide that ends at a height of 1.80 m above the pool. If a child starts from rest at po

int A and lands in the water at point B, a horizontal distance L = 2.42 m from the base of the slide, determine the height h of the water slide.
Physics
1 answer:
irina [24]3 years ago
4 0

Answer:0.813 m

Explanation:

Given

height at which slide ends is h=1.8 m

Horizontal distance L=2.42 m moved by child from slide base to pool

At slide end Child has only horizontal velocity

horizontal velocity is given by u=\sqrt{2gH}

where H=height of slide

time taken by child to complete 1.8 m height

t=\sqrt{\frac{2h}{g}}

t=\sqrt{\frac{2\times 1.8}{9.8}}

t=0.606 s

Horizontal distance moved in this time

L=2.42=u\times t

2.42=\sqrt{2gH}\times 0.606

\sqrt{2gH}=3.99

H=0.813 m      

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Why earth have gravity​
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You have a chemical sealed in a glass container filled with air. The glass container has a mass of 200.0 grams and the chemical
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Explanation:

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what is the total mass after chemical burn

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3 0
4 years ago
Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
Kipish [7]

Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

6 0
3 years ago
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