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katen-ka-za [31]
3 years ago
11

A long thing bar of copper is heated evenly along it's length

Physics
1 answer:
leva [86]3 years ago
8 0

Answer:

It becomes less heavy

Explanation:

Hopefully it helps!

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A 320-g ball and a 400-g ball are attached to the two ends of a string that goes over a pulley with a radius of 8.7 cm. Because
Gnom [1K]

To solve this problem, it is necessary to apply the concepts related to force described in Newton's second law, so that

F = ma

Where,

m = mass

a = Acceleration (Gravitational acceleration when there is action over the object of the earth)

Torque, as we know, is the force applied at a certain distance, that is,

\tau = F*d

Where

F= Force

d = Distance

Our values are given as,

m_1 = 0.32Kg

m_2 = 0.4Kg

d = 8.7*10^{-2}m

Since the system is in equilibrium the difference of the torques is the result of the total Torque applied, that is to say

\tau = T_2-T_1

\tau = F_2*d-F_1*d

\tau = m_2g*d-m_1*g*d

\tau = (m_2-m_1)g*d

\tau = (0.4-0.32)(9.8)(8.7*10^{-2})

\tau = 0.068N\cdot m

Therefore the magnitude of the frictional torque at the axle of the pulley if the system remains at rest when the balls are released is \tau = 0.068N\cdot m

8 0
3 years ago
Paul sprinkles iron filings onto a piece of paper
Ksivusya [100]

Answer:

The other one

Explanation:

6 0
3 years ago
(Write the answer in fair test) You have to investigate whether surface area affects how fast some water evaporates
Damm [24]

Answer:

Evaporation increases with an increase in the surface area

If the surface area is increased, then the amount is of liquid that is exposed to air is larger. More molecules can escape with a wider surface area.

Explanation:

7 0
3 years ago
The speed of an object increases by 30 m/s in 10 seconds. what is its acceleration?
Sladkaya [172]

Hope this helps you

4 0
3 years ago
Consider the power dissipated by the two circuits in the video. The ratio of power dissipation in the parallel circuit to that i
klemol [59]

Answer:

A)

Explanation:

  • In a series circuit,  if we assume that each bulb has a resistance R, the equivalent resistance, is the sum of all the resistances, in this case, 3*R.
  • The power dissipated in this resistance, can be expressed as follows:

       P_{series}  = V*I = V*(\frac{V}{R_{eq}})  = \frac{V^{2}}{3*R} (1)

  • In the parallel circuit, the equivalent resistance, is equal to 1/3*R.
  • The power dissipated in this resistance, can be expressed as follows:

       P_{//}  = V*I = V*(\frac{V}{R_{eq}})  = \frac{V^{2}}{\frac{R}{3} } (2)

  • In order to take the ratio of both powers, we can divide both sides of (1) and (2) , as follows:

        \frac{P_{//} }{P_{series} } =\frac{V^{2}}{\frac{R}{3}} * \frac{3*R}{V^{2}} = 9

  • The ratio of power dissipation in the parallel circuit to that in the series circuit is 9 times.
5 0
4 years ago
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