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ICE Princess25 [194]
3 years ago
11

A student applies a 10 N force to a wood block with a mass of 5 kg. The block is pushed across four different surfaces. The acce

lerations of the block are recorded. Which surface showed the least friction?
Physics
1 answer:
Furkat [3]3 years ago
8 0

The complete question is as follows: A student is subjected to a reaction force of 10 N northward from a 5 kg block while pushing the block over a smooth, level surface. Ignoring friction, what is the acceleration of the block?

Answer: The acceleration of the block is 2 m/s^{2}.

Explanation:

Given: Force = 10 N

Mass = 5 kg

It is known that force applied on an object is the product of mass and acceleration.

Mathematically, F = m \times a

where,

F = force

m = mass

a = acceleration

Substitute the values into above formula as follows.

F = m \times a\\10 N = 5 kg \times a\\a = \frac{10}{5}\\= 2 m/s^{2}

Thus, we can conclude that the acceleration of block is 2 m/s^{2}.

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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
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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}}

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