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abruzzese [7]
3 years ago
13

A 120 -kg crate accelerates toward the positive x-direction . If the magnitude of the force due to friction is 74.4 N , what for

ce must pull on the crate to give it a 2.5 m/s ^ s acceleration ?
Physics
1 answer:
Usimov [2.4K]3 years ago
4 0

Answer:

The applied force to accelerate the crate is 374.4 N

Explanation:

Given;

mass of the crate, m = 120 kg

magnitude of force due to friction, Fk = 74.4N

acceleration of the crate, a = 2.5 m/s²

The net horizontal force on the crate is calculated as;

∑Fx = F - Fk

ma = F - Fk

F = ma + Fk

where;

F is the applied force to accelerate the crate by 2.5 m/s²

F = (120 x 2.5) + (74.4 N)

F = 300 N + 74.4 N

F = 374.4 N

Therefore, the applied force to accelerate the crate is 374.4 N

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Which statement describes a controlled experiment?
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7 0
3 years ago
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cable
kompoz [17]

Answer:

4.44s

Explanation:

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables

since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation

T=2π√L/g

g=acceleration due to gravity which is 9.81m/s2

the length of the supporting cable is 4.9m

T the period

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4.44s

5 0
3 years ago
Please help me this is timed . <br> Find x if a = 3.0 m/s^2
storchak [24]

Answer:

x = 50 N

Explanation:

Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.

Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.

_______

F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a

m (mass which we are given) = 150 kg

a (acceleration which we are given) = 3.0m/s

________

So F = m × a → F2 - F1 = m × a →

500 - F1 = 150 × 3.0 → 500 - F1 = 450 →

-F1 = -50 → F1 = 50

5 0
3 years ago
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