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3 years ago

How does increasing the distance between charged objects affect the electric force between them?

Physics

1 answer:

mina [271]3 years ago

6 0

Answer: energy force

Explanation:

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A 40 kg skateboarder runs at a constant velocity of 12 m/s and jumps off a stationary skateboard with a mass of 8 kg. What is th

horsena [70]

Answer:

60 m/s (in the opposite direction)

Explanation:

By conservation of momentum,

Initial momentum of board + Initial momentum of man = Final momentum of board + Final momentum of man

=> 0 + 40*12 = 8*(-v) + 0

=> v = -60 [ Negative indicates opposite direction ]

7 0

4 years ago

PLZ HELP, GIVING BRAINLIEST!!!

Y_Kistochka [10]

A snow blower is more powerful because it's easier, and it moves large amounts of snow easily. Shoveling uses all of <em>your power.</em> It can be difficult and it takes a long time to get the snow off of your lawn.

Therefore, using a snow blower is more powerful than using a shovel.

I hope this helped!

6 0

3 years ago

What is the current produced by the solar cells of a pocket calculator through which 4.00 C of charge passes in 4.00 hr? Give yo

AfilCa [17]

The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

Q = 4.00C

Δt = 4.00hr = 14400s

Plug in the values and solve for I:

I = 4.00/14400

I = 0.000277777778A

I = 0.278mA

3 0

3 years ago

What type of erosion formed the grand canyon??

Semmy [17]

the correct answer is physical erosion. hope it helps

3 0

3 years ago

The angular momentum of a flywheel having a rotational inertia of 0.200 kg · m2 about its central axis decreases from 3.80 to 0.

Bingel [31]

Answer:

14.3065 rad

-36.777 J

-28.29 W

Explanation:

$L_f$ = Final angular momentum = 0.6 kgm²/s

$L_i$ = Initial angular momentum = 3 kgm²/s

I = Moment of inertia = 0.2 kgm²

Torque is given by

$\tau=\dfrac{L_f-L_i}{t}\\\Rightarrow \tau=\frac{0.6-3.8}{1.3}\\\Rightarrow \tau=-2.46\ Nm$

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2$

Initial angular speed is given by

$\omega_i=\dfrac{L_i}{I}$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}$

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=\dfrac{L_it+\dfrac{1}{2}\tau t^2}{I}\\\Rightarrow \theta=\dfrac{3.8\times 1.3+\dfrac{1}{2}\times -2.46\times 1.3^2}{0.2}\\\Rightarrow \theta=14.3065\ rad$

The angle is 14.3065 rad

Work done is given by

$W=\tau \theta\\\Rightarrow W=-2.46\times 14.95\\\Rightarrow W=-36.777\ J$

The work done on the wheel is -36.777 J

Power is given by

$P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{-36.777}{1.3}\\\Rightarrow P=-28.29\ W$

The power is -28.29 W

6 0

3 years ago