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3 years ago

An object completes 10 cycles in 50s.

Physics

2 answers:

Mashcka [7]3 years ago

8 0

Answer:

Explanation:

nataly862011 [7]3 years ago

6 0

Explanation:

Given parameters:

Number of cycles = 10cycles

Time taken = 50s

Unknown:

Period of its rotation = ?

Frequency of its rotation = ?

Solution:

The period of rotation is the time taken time taken for a body to complete a number of rotation;

Period = $\frac{t}{n}$

t is the time

n is the number of cycles

Period = $\frac{50}{10}$ = 5s

The period is 5s

Now, frequency is the inverse of period;

F = $\frac{1}{period}$ = $\frac{1}{5}$ = 0.2Hz

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andriy [413]

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3 years ago

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3 years ago

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 .

Elis [28]

<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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3 years ago

A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de

EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/π $r^{2}$

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / π $r^{2}$ ex

E = $\frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6} } = 17.368 * 10^{9} Pa = 17.4 GPa$

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = $\frac{1}{E}$ (бy - v (бx + бz)) = $-\frac{v}{E}$ бx

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Finally

ey = Δr / r

Δr = ey * r = 10 * $-0.63* 10^{-6} mm = -6.3 * 10^{-6} mm$

Δd = 2Δr = $-12.6 * 10^{-6} mm$

Explanation:

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Tamiku [17]

Answer:

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Explanation:

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3 years ago