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Sedaia [141]
3 years ago
7

18 kilogram Mass Blokus addressed a level surface if the coefficient of static friction between the Block in the surface is 0.6

what is horizontal force is required to take gravity 10m/s2
​
Physics
1 answer:
Tasya [4]3 years ago
4 0

Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what  horizontal force is required to just move the blcok ( take gravity as 10m/s2 )

Answer:

108 N

Explanation:

From the question,

Applying

F' = mgμ................ Equation 1

Where F' = Frictional force = horizontal  force required to just move the block,  m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.

From the question,

Given: m = 18 kg, μ = 0.6, g = 10 m/s²

Substitute these values into equation 1

F' = 18×0.6×10

F' = 108 N

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The wavelength in vacuum is equal to 428.8 nm.

Explanation:

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According to Newton's Second Law, the force of the club hitting the golf ball will cause it to accelerate. At the moment of impa
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Answer:

Option B

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<h3>According to Newton's third law, for every reaction there will be equal and opposite reaction</h3>

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8 0
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A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total
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Answer:

<em>1) 14 kW</em>

<em>2) 4.67 x 10^-5 N</em>

Explanation:

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Intensity of sun's radiation incident on earth = 1.4 kW/m^2

Solar power absorbed = ?

We know that the intensity of radiation on a given area is

I = \frac{P}{A}

where I is the intensity of the radiation

P is the power absorbed due to this intensity on a given area

A is the area on which this radiation is incident

From the equation, we have

P = IA

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p = I/c

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c is the speed of light = 3 x 10^8 m/s

substituting values, we have

p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa

we know that Force = pressure x area

therefore force on the solar panels is

F = 4.67 x 10^-6 x 10 = <em>4.67 x 10^-5 N</em>

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