3 years ago

Part C

Physics

1 answer:

iVinArrow [24]3 years ago

7 0

Answer:the wave lengths are shorter and faster

Explanation:

Plato

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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:

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

6 0

3 years ago

If you wanna make it easier you can number them going down 1,2,3, or 4 & just tell me the number

prisoha [69]

the number is 1 I hope this help

8 0

3 years ago

A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy. The energy used to do work

nika2105 [10]

Answer:

480J

Explanation:

Using the formula:

Delta U = Q - W

Q:Heat (J)

Delta U: Changes in internal Energy (J)

W:Work (J)

We can plug in the give numbers, Q and W.

Delta U = 658J - 178J = 480J

6 0

3 years ago

If we want to apply Gauss's Law on an infinite length line of charges, how much open surfaces we can obtain in the closed cylind

GrogVix [38]

Answer:

Explanation:

There are two open surfaces obtained in a closed cylindrical Gaussian surface when we apply Gauss's Law on an infinite length line of charges. The two surfaces are the upper circular base and the lower circular base.

Moreover, Gauss Law states that the net electric flow outside to a closed surface is proportional to the electric charge within the closed surface.

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3 years ago

An elephant and a mouse when dropped from the same height (ignoring

attashe74 [19]

Answer:

True

Explanation:

5 0

3 years ago