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Harman [31]
1 year ago
8

calculate the weight of an object of mass 100 kg on the surface of a Planet whose mass and diameter are 4.8*10^24 kg and 12000km

respectively.​
Physics
1 answer:
PilotLPTM [1.2K]1 year ago
6 0

The weight of an object mass of 100 kg on the surface of a Planet willl is 981 N.

<h3>What is weight?</h3>

The weight of matter is found as the product of the mass and the gravitational acceleration;

Given data;

Weight of an object,W =?

Mass,m = 100 kg

\rm W= mg \\\\ \rm W= 100  \ kg  \times 9.81 \ m/sec^2 \\\\ W=981 \ N

Hence the weight of an object willl be 981 N.

To learn more about the weight refer;

brainly.com/question/10069252

#SPJ1

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1/3 the weight than it is on earth

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Myopia is a condition of the eye where the crystalline lens focuses the light rays to a position between the lens and the retina
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Answer: concave lens

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2 years ago
How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?
lesantik [10]

Answer:

Power = 70 W

Explanation:

Given that,

Force, F = 70 N

Height, h = 5 m

Time, t = 5 s

We need to find the power of the object. We know that,

Power = work done/time

Put all the values,

P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W

So, the required power is 70 W.

3 0
2 years ago
A street light is on top of a 8 foot pole. Joe,
zubka84 [21]
8/4 = y/y-x

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Hope this helps
5 0
3 years ago
A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.
enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, KE_{0}  = 0.5 mu^{2}

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, KE_{f} = 0.5mv^{2}

a) Using the principle of energy conservation,

KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0
3 years ago
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