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3 years ago

How is energy transferred whenhitting a nail?

Physics

1 answer:

Juli2301 [7.4K]3 years ago

7 0

Answer:

kinetic energy

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Answer:

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3 years ago

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A has the magnitude 14.4 m and is angled 51.6° counterclockwise from the positive direction of the x axis of an xy coordinate sy

Ad libitum [116K]

Answer:

Ã in unit vector notation = 12.26485i + 7.54539j

B in unit vector notation = 16.3516i + 3.11529j

Explanation:

The detailed steps and calculation is shown in the attachment.

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3 years ago

Two identical conducting spheres, A and B, carry equal charge. They are stationary and are separated by a distance much larger t

podryga [215]

Answer:

8F_i = 3F_f

Explanation:

When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.

Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.

Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.

The electrostatic force, Fi, in the initial configuration can be calculated as follows.

$F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f$

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3 years ago

What needed to be present in marine mud to form fossils fuels?

garri49 [273]

It would be oraganic matter I think.

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3 years ago

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g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl

Fynjy0 [20]

Answer:

$v=1.5m/s$

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write $mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}$ . Since $v=r\omega$ (the ball rolls without slipping) and for a solid sphere $I=\frac{2mr^2}{5}$ , we have:

$mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}$

So our translational speed will be:

$v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s$

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3 years ago

How is energy transferred whenhitting a nail?​

How is energy transferred whenhitting a nail?