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kramer
3 years ago
13

A 60 kg student is standing atop a spring in an elevator that is accelerating upward at 3.0 m/s2. The spring constant is 2.5 x 1

03 N/m. By how much is the spring compressed?
Physics
1 answer:
dimaraw [331]3 years ago
6 0

the spring will be compressed by 0.3072 m

Explanation:

acceleration of elevator=3 m/s²

mass of student= 60 Kg

spring constant=2.5 x 10³ N/m

the force on the student is given by F = m ( g +a)

F=60 (9.8+3)

F=768 N

now the formula for spring force is given by

F= k x

768= 2.5 x 10³ (x)

x=0.3072 m

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Explanation:

Given

Power\left ( P\right )=150 MW

mass of core\left ( m\right )=1.60\times 10^5 kg

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Now

P=mc\frac{\mathrm{d}T}{\mathrm{d} t}

150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}

Thus \frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}

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6 0
3 years ago
A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni
mina [271]

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Induced emf, \epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}

Explanation:

The varying magnetic field with time t is given by according to equation as :

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So, the induced emf in the loop as a function of time is A\dfrac{B_{max}e^{-t/\tau}}{\tau}. Hence, this is the required solution.

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3 years ago
Please help!! I will give brainliest!
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Answer:

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So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0
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