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Dmitry [639]
3 years ago
9

If the weight of an object on the moon is 1/6 of the weight of an object on earth. What

Physics
2 answers:
zubka84 [21]3 years ago
8 0

Answer:

1200N

Explanation:

Suppose a body of mass "m" and its weight on the moon is Wm (where W is the weight and "m" is the moon ;which means weight on the moon).Mass of the moon is "M"

and its radius is "R"

Weight of an object on the moon = "F"(Force)with which the moon pulls.

Wm = GM*m/r2

Weight of the same object on the earth is We(where W is the weight and "e" is the earth; which means weight on the earth).

Mass of the earth is 100 times of that of the moon.

Radius of the moon = R

Radius of the Earth = 4R

Weight of the object on the moon =

We = G100M*m/(4R)2(Pronounced 4 R square)

We = G100M*m/(16R)2(Pronounced 16 R square)

Wm/We = G * M * m * 16R2/R2 * g * 100M * m

=16/100

Therefore, 4800N on earth= 1200N on moon

tatuchka [14]3 years ago
5 0

Answer:

1200 N

Explanation:

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Options

B,D,E

From the question we are told

Choose the correct definitions of speed, velocity, and acceleration. Check all that apply.

  • Acceleration tells us in which direction the object is going. Speed is the rate of change of velocity in time.
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Speed is used to describe how fast the object is moving and tells us in which direction it is going.

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3 0
3 years ago
A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational
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<span> gravitational force varies based on 1/r^2
when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.
</span><span>As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)
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5 0
4 years ago
Read 2 more answers
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

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Pressure is measured as force per unit area, which is the third option.

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Pressure is amount of force that is applied over a given area divided by the size of this area.

Pressure is calculated by multiplying the force applied on the object by the area covered.

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Answer:

Explanation:

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F=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{\left (911\times 10^{-9}  \right )^2}

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4 0
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