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pishuonlain [190]
3 years ago
10

Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph .

Normally Richard drives at the speed limit, but today he is running late and decides to take his chances by driving at 73 mph .
Physics
1 answer:
Elis [28]3 years ago
5 0
<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

v=\frac{d}{t} \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_{1}=\frac{d}{v_{1}} \\ \\ t=\frac{135}{65} \\ \\ t_{1}=2.07 \ hours

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

t_{2}=\frac{135}{73} \\ \\ t_{2}=1.85 \ hours

So he saves the time t_{s}:

t_{s}=t_{1}-t_{2}=2.07-1.85=0.22 \ hours

In minutes:

t_{s}=0.22h\left(\frac{60min}{1h}\right) \\ \\ \boxed{t_{s}=13.2min}

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A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu
Ghella [55]

The angular frequency of the cyclotron is 0.07 x 10^{8}Hz.

<h3>What is angular frequency?</h3>
  • Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
  • It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.
<h3>What is cyclotron?</h3>

The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.

<h3>Calculation of angular frequency:</h3>

Given,

B = 0.47 T

r = 0.68

mass of proton = 1.6x10^{-27}

q = 1.6 x 10^{-19}

so, the frequency is:

f = qB/2\pim

f = 1.6 x 10^{-19} x 0.47/2x3.14x1.6x10^{-27}

f = 0.07 x 10^{8}

Hence, the angular frequency of the cyclotron is  0.07 x 10^{8}Hz.

Learn more about angular frequency here:

brainly.com/question/14244057

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