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4 years ago

An asteroid has a radius of 110,000 m

Physics

1 answer:

nydimaria [60]4 years ago

7 0

Answer: M = 6.13 × 10^18 kg

Explanation:

g = GM/r2,

Where

The mass M of the asteroid = ?

The radius r = 110000 m

g = 0.0338 m/s^2

G is the gravitational constant.

SI units its value is approximately 6.674×10^−11m3⋅kg−1⋅s−2

Using the formula

g = GM/r2

Cross multiply

GM = gr^2

6.674×10^-11M = 0.0338 × 110000^2

M = 408×10^6/6.674×10^-11

M = 6.13 × 10^18 kg

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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

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

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

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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3 years ago

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Assuming the gas behaves ideally,
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konstantin123 [22]

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3 years ago

A car drives at steady speed around a perfectly circular track.

gayaneshka [121]

Answer:

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4 years ago