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3 years ago

(give the answer or I'll report) please solve it with the steps

Physics

1 answer:

Leona [35]3 years ago

3 0

Answer:

The required elastic potential energy is 0.068J

Explanation:

Given,

Spring force constant, k= 3.4 N/M

Spring stretch length/ Displacement, x= 0.2m

We know,

Elastic potential energy, V= $\frac{1}{2} kx^{2}$

= $\frac{1}{2}*3.4*(0.2)^{2}$

= $\frac{0.136}{2}$

=0.068

∴The required elastic potential energy is 0.068J

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You rub a clear plastic pen with wool, and observe that a strip of invisible tape is attacted to the pen. Assuming that the pen

gayaneshka [121]

Answer:

option B

Explanation:

When you rub a clear plastic pen with the wool the plastic pen gets charges this phenomenon is known as frictional charging.

Due to rubbing, the pen gets negatively charged.

We know, opposite charge attract each other and the same charge repel each other.

So, when the pen is negatively charged the tape might be positively charged or the tape might be uncharged.

Hence, the correct answer is option B

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3 years ago

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Vinvika [58]

Answer:

volume;mass

Explanation:

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3 years ago

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2 years ago

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A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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(give the answer or I'll report) please solve it with the steps​

(give the answer or I'll report) please solve it with the steps