Change to Kelvin 60 degree Celicius؟

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

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

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

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

2 years ago

PLEASE HELP ASAP!!

kolezko [41]

A. Wedge

and

B.Screw

A screw is like an inclined plane wrapped around a pole or rod.

and a wedge looks kind of looks like this / and is inclined.

Hope this helps!

Please mark me brainliest! Thanks!

5 0

2 years ago

A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.4 m/s. The drag force is of the form bv^2 What is the value of b?

julia-pushkina [17]

Drag Force = bv^2 = ma; a = g = 9.81 m/s^2

b = mg/v^2 = (0.0023×9.81)/(9.4^2)

b = 0.000255

6 0

2 years ago

Anybody help me, please?<br><br> Describe one periodic trend that you learned about.

Korolek [52]

Answer:

the renegade

Explanation: charklie dfamielo

8 0

1 year ago

A 65.0-kg runner has a speed of 5.20 m/s at one instant dur- ing a long-distance event. (a) What is the runner’s kinetic energy

vladimir2022 [97]

Answer:

a)KE=878.8 J

b)W=2636.4 J

Explanation:

Given that

mass ,m = 65 kg

Initial speed ,u = 5.2 m/s

a)

We know that kinetic energy KE is given as follows

$KE=\dfrac{1}{2}mu^2$

m=mass

u=velocity

Now by putting the values in the above equation we get

$KE=\dfrac{1}{2}\times 65\times 5.2^2\ J$

KE=878.8 J

b)

We know that

Work done by all forces = Change in the kinetic energy

The final velocity , v= 2 u = 2 x 5.2 m/s

v= 10.4 m/s

$W=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2$

Now by putting the values in the above equation we get

$W=\dfrac{1}{2}\times 65\times 10.4^2-\dfrac{1}{2}\times 65\times 5.2^2\ J$

W=2636.4 J

a)KE=878.8 J

b)W=2636.4 J

8 0

2 years ago