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alekssr [168]
2 years ago
13

A turntable is designed to acquire an angular velocity of 32.4 rev/s in 0.5 s, starting from rest.

Physics
2 answers:
frozen [14]2 years ago
6 0

Answer:

not sure lol

Explanation:

ummmmmm look it up on peersanswer.com

horrorfan [7]2 years ago
3 0
Well, the acceleration is the difference of speeds divided by the time period. \frac{32.4}{0.5}=64.8rev/s^2.
One rev/s is 2\pi rad/s^2, so our final result is 64.8*2\pi=407.15rad/s^2.
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Convert the following angles in degrees to radians:<br><br>(a) 300°<br>(b) 18°<br>(c) 105°​
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A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
PLEASE HELP
Arisa [49]

Answer:

C should be the correct one

5 0
3 years ago
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