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4 years ago

A spring of original length 3.0 cm is extended to a total length of 5.0 cm by a force of 8.0N

Physics

2 answers:

miss Akunina [59]4 years ago

8 0

Answer:

12.0N

Explanation:

This problem is solved using Hooke's law which states that the extension or compression experienced by an elastic material is proportional the the force or load applied provided that the elastic limit is not exceeded.

$F=ke..................(1)$

where F is the force applied, k is the elastic constant and e is the extension of the material.

$e=l_1-l_o..................(2)$

were $l_o$ is the original length of the material and $l_1$ is its total length after extension.

Given;

F = 8.0N

$l_o=3cm\\l_1=5cm\\k=?$

therefore,

$e=5-3\\e=2cm$

substituting all necessary values into equation (1), we get;

$8=k*2\\k=\frac{8}{2}\\k=4N/cm$

The force required to give the spring a total length of 6cm is thus calculated;

in this case,

$l_1=6cm\\therefore\\e=6-3\\e=3cm$

hence F is given as follows;

$F=(4N/cm) *3cm\\F=12.0N$

hjlf4 years ago

6 0

4848484 5 5 5 5 5 =++++

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A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th

Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

$E=-\frac{d}{dt}(BACOS\alpha )\\$

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using $\pi r^{2} \\$ where r is the radius of 5cm or 0.05m

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since we no that the angle is at $0^{0}$

we determine the magnitude of the magnetic filed

$B=0.5e^{-t} \\t=2s$

$E=-(0.5e^{-2} * 0.00785)$

$E=-0.000532v\\$

the Magnitude of the voltage is 0.000532V

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$V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A$

$I=2.6mA$

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3 years ago

A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis

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Answer:

2 rad/s

Explanation:

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Answer:

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6 0

3 years ago

!!!!!!!!!!!PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!

Alexeev081 [22]

You might want to do a re-erp on this equation

5 0

3 years ago

Read 2 more answers