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aev [14]
3 years ago
6

Find the force, in N that does 0.0284 kilojoules of work in moving a book a distance of 4.00 meters

Physics
1 answer:
Anarel [89]3 years ago
3 0
Work equals force times distance. When you move an object, you are exerting a force onto it. By exerting a force on the object, you are actually displacing it from its initial position. You cannot apply force to the object without altering its position. Keep in mind that when you exert work, you are exerting energy too. Note that 1 kilojoule = 1000 joules and joules is equal to Newton meter. So,  

W = F*d
F = W/d
F = (0.0284 kJ)(1000J/1kJ)(1N-m/1J)/4m
<u>F =  7.1 N</u>
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8 0
3 years ago
#2 - the diagram<br> Thank you!!!! :D
s344n2d4d5 [400]

Answer:

answer in the explanation!!

Explanation:

1: light being emitted from a light bulb is called electromagnetic waves ; transverse wave.

2: sound coming from a violin is a mechanical wave ; longitudinal wave.

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8 0
3 years ago
Read 2 more answers
Fill in the blank
nignag [31]

Answer:

temperature

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5 0
3 years ago
5. A cable is attached 32.0 m from the base of a flagpole that is about to
soldi70 [24.7K]

Answer:

The length of the flagpole is approximately 87.43 m

Explanation:

The given parameters of the cable attached to the flagpole are;

The point along the flagpole at which the cable is attached = 32.0 m

The angle with respect to the ground at which the raising of the flagpole is halted = 60.0°

The downward force exerted by the cable, F_v = 1.233 × 10⁴ N

The force exerted by the cable to the left = 1.233 × 10⁴ N

Let 'W' represent the weight of the flagpole, at equilibrium, we have;

The sum of vertical forces = 0

Therefore;

F_v + W - R = 0

W - R = -1.233 × 10⁴ N

Taking moment about the support at the base of the pole, we get;

1.233 × 10⁴ × d × cos(60.0°) - 1.233 × 10⁴ ×d× sin(60.0°) + W × d/2 ×cos(60.0°) = 0

∴ W × d/2 ×cos(60.0°) ≈  4513.093·d  

W = 2 × 4513.093/(cos(60.0°)) ≈ 18,052.373 N

R = 18,052.373 + 1.233 × 10⁴ ≈ 30,382.373

R ≈ 30,382.373 N

Taking moment about the point of attachment of the cable to the ground, we have;

W × ((d/2) × cos(60.0°) + 32) = R × 32

∴ (d/2) = ((30,382.373 × 32/18,052.373) - 32)/(cos(60.0°)) ≈ 43.71281

d = 2 × 43.71281 ≈ 87.43

The length of the flagpole, d ≈ 87.43 m

7 0
3 years ago
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