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creativ13 [48]
4 years ago
12

Suppose you do work on an object from rest to a certain speed v. Why does it take four times the amount of work to increase its

speed from rest to a speed of 2v?
Physics
1 answer:
Otrada [13]4 years ago
8 0
Answer: it is due to friction.

Explanation: if you were to push an object that is at rest, there is static friction that you would first need to overcome. An example would be pushing a car as it is in neutral. Pushing it takes more force and work to overcome that friction.
But once it overcomes static, it will then turn to kinetic friction. Once the car starts to move, you notice that it is easier to push. This is due to the fact that kinetic friction would equal normal force times the constant while the static friction is greater than or equal to the normal force times the constant.

Hope this helps!!!
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8 0
4 years ago
Need help on this thermal physics question ASAP please
Trava [24]
Since V is 5 times larger than C

A) 0V = 25C

25C to 0C = -25
-25 / 5 = -5

so 0C = -5 V

B)

20V x 5 = 100 C


6 0
3 years ago
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lisabon 2012 [21]
B 200N; 200N/10kg=20m/s2
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The voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery wears out. At the same time, the resi
Anastasy [175]

Answer:

-0.5\times 10^{-4} A/s

Explanation:

We are given that

\frac{dV}{dt}=-0.01 V/s

R=600 ohms

I=0.04 A

\frac{dR}{dt}=0.5ohm/s

V=IR

\frac{dV}{dt}=\frac{\partial V}{dI}\frac{dI}{dt}+\frac{\partial V}{dR}\frac{dR}{dt}

\frac{dV}{dt}=R\frac{dI}{dt}+I\frac{dR}{dt}

Substitute the values

-0.01=600\times \frac{dI}{dt}+0.04\times 0.5

-0.01-0.04\times 0.5=600\frac{dI}{dt}

-0.03=600\frac{dI}{dt}

\frac{dI}{dt}=\frac{-0.03}{600}=-0.5\times 10^{-4}A/s

5 0
4 years ago
As a capacitor is being charged in an RC circuit. the current flowing through a resistor isa) increasingb) decreasingc) constant
Montano1993 [528]
<h2>Answer: decreasing</h2>

An RC circuit is an electrical circuit composed of resistors and capacitors, where the charging time T of the circuit is proportional to the magnitude of the electrical resistance R and the capacity C of the capacitor.  

As shown below:

T=R.C

In this context, the electrical resistance is the opposition to the flow of electrons when moving through a conductor.

Therefore:

<h2>When a capacitor is being charged in an RC circuit, the current flowing through a resistor <u>decreases</u>.</h2>

And the correct option is b.

4 0
3 years ago
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