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Vanyuwa [196]
3 years ago
6

A 5.0kg toolbox is raised from the ground by a rope. If the upward acceleration of the bucket is 2.5 m/s^2, find the force exert

ed by the rope on the toolbox. Acceleration due to gravity is 9.8m/s^2.
Physics
1 answer:
swat323 years ago
8 0

Answer:

62 N

Explanation:

Sum of the forces on the toolbox:

∑F = ma

T − mg = ma

T = mg + ma

T = m (g + a)

T = (5.0 kg) (9.8 m/s² + 2.5 m/s²)

T = 61.5 N

Rounded to two significant figures, the force exerted by the rope is 62 N.

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The combination of all forces that act on an object are (2 points)
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Answer:

A) The net force

Explanation:

If two forces of equal strength act on an object in opposite directions, the forces will cancel, resulting in a net force of zero and no movement.

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You have a stopped pipe of adjustable length close to a taut 62.0 cmcm, 7.25 gg wire under a tension of 4710 NN. You want to adj
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Answer:

6 cm long

Explanation:

F = 4110N

Vo(speed of sound) = 344m/s

Mass = 7.25g = 0.00725kg

L = 62.0cm = 0.62m

Speed of a wave in string is

V = √(F / μ)

V = speed of the wave

F = force of tension acting on the string

μ = mass per unit density

F(n) = n (v / 2L)

L = string length

μ = mass / length

μ = 0.00725 / 0.62

μ = 0.0116 ≅ 0.0117kg/m

V = √(F / μ)

V = √(4110 / 0.0117)

v = 592.69m/s

Second overtone n = 3 since it's the third harmonic

F(n) = n * (v / 2L)

F₃ = 3 * [592.69 / (2 * 0.62)

F₃ = 1778.07 / 1.24 = 1433.927Hz

The frequency for standing wave in a stopped pipe

f = n (v / 4L)

Since it's the first fundamental, n = 1

1433.93 = 344 / 4L

4L = 344 / 1433.93

4L = 0.2399

L = 0.0599

L = 0.06cm

L = 6cm

The pipe should be 6 cm long

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