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3 years ago

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A block weighs 15 n and is suspended from a spring that is attached to the ceiling. the spring stretches by 0.075 m from its uns

trained length. by how much does the spring stretch when a 33-n block is suspended from it?

1 answer:

Illusion [34]3 years ago

4 0

We can salve the problem by using the formula:

$F=kx$

where F is the force applied, k is the spring constant and x is the stretching of the spring.

From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:

$k=\frac{F}{x}=\frac{15 N}{0.075 m}=200 N/m$

By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied:

$x=\frac{F}{k}=\frac{33 N}{200 N/m}=0.165 m$

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Explanation:

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$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$

$F_2=\dfrac{F_1}{A_1}\times A_2$

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3 years ago

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2 years ago

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Explanation:

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$v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}$

$v=5.92\times 10^7 i\ m/s$ (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

$F=q(v\times B)$

$F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})$

Since, $i\times j=k\ \\j\times k=i\\k\times i=j$

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$F=4.95\times 10^{-16}\ N$

(b) Let a is the acceleration of the electron. It can be calculated as :

$a=\dfrac{F}{m}$

$a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}$

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Answer:

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