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fredd [130]
3 years ago
14

A wood block is sliding up a wood ramp. if the ramp is very steep, the block will reverse direction at its highest point and sli

de back down. if the ramp is shallow, the block will stop when it reaches its highest point.
Physics
1 answer:
DanielleElmas [232]3 years ago
5 0
<span>Answer: So it gets to the top of the ramp and stops. The parallel force pushing it down the ramp is mg sin θ, but for it to move, the frictional force must be overcome. This frictional force is μmg cos θ, where μ is the coefficient of static friction. For movement, then, mg sin θ > μmg cos θ ==> tan θ > μ ==> θ > arctan 0.5 = 26.565° ==> θ = 27°</span>
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Dahasolnce [82]

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

20N/4 = 5N

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3 years ago
Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
Fudgin [204]

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
  • g'=1.63\ m.s^{-2}

<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

  • G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}
  • m=7.35\times 10^{22}\ kg
  • r=1.74\times 10^6\ m
  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

4 0
3 years ago
If the work done on a object is 100 J and it comes to a rest on a surface with a mass of 10
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During lightning strikes from a cloud to the ground, currents as high as 2.50×10^4 Amps can occur and last for about 40.0 micros
dangina [55]

Answer:

1 C

Explanation:

The intensity of electric current is defined as

I=\frac{q}{t}

where

I is the current

q is the amount of charge transferred

t is the time interval during which the charge is transferred

For the lightning in this problem, we have

I=2.50\cdot 10^4 A is the current

t=40.0 \mu s = 40.0\cdot 10^{-6} s is the time interval

Solving the formula for q, we find the amount of charge transferred:

q=I t = (2.50\cdot 10^4 A)(40.0\cdot 10^{-6}s)=1 C

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