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GarryVolchara [31]
3 years ago
5

A frictionless inclined plane is 8.0 m long and rests on a wall that is 2.0 m high. How much force is needed to push a block of

ice weighing 300.0 N up the plane

Physics
1 answer:
adoni [48]3 years ago
7 0

Given Information:  

Inclined plane length = 8 m

Inclined plane height = 2 m

Weight of ice block = 300 N

Required Information:  

Force required to push ice block = F = ?  

Answer:  

Force required to push ice block =  75 N

Explanation:  

The force required to push this block of ice on a inclined plane is given by

F = Wsinθ

Where W is the weight of the ice block and θ is the angle as shown in the attached image.

Recall from trigonometry ratios,

sinθ = opposite/hypotenuse

Where opposite is height of the inclined plane and hypotenuse in the length of the inclined plane.

sinθ = 2/8

θ = sin⁻¹(2/8)

θ = 14.48°

F = 300*sin(14.48)

F = 75 N

Therefore, a force of 75 N is required to push this ice block on the given inclined plane.

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Answer:

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3 years ago
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Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

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On substituting the values into the above equation, we get,

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1.67x = 1+x

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Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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