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Doss [256]
3 years ago
11

A 50 kg box hangs from a rope. What is the tension in the rope if: The box is at rest? The box moves up at a steady 5.0 m/s? The

box has vy=5.0 m/s and is speeding up at 5.0 m/s2? The box has vy=5.0 m/s and is slowing down at 5.0 m/s2?
Physics
1 answer:
Verizon [17]3 years ago
6 0

Answer:

(a) T_{1}=490N

(b) T_{2}=240N

Explanation:

For Part (a)

Given data

The box moves up at steady 5.0 m/s

The mas of box is 50 kg

As ∑Fy=T₁ - mg=0

T_{1}=mg\\T_{1}=(50kg)(9.8m/s^{2} ) \\T_{1}=490N

For Part(b)

Given data

v_{iy}=5m/s\\ a_{y}=-5.0m/s^{2}

As ∑Fy=T₂ - mg=ma

T_{2}=mg+ma_{y}\\T_{2}=m(g+ a_{y})\\T_{2}=50kg(9.8-5.0) \\T_{2}=240N

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Now, if the charge q is mid way between the field, then distance is \frac{d}{2}.

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\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}

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Answer:

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Explanation:

Given;

mass of bullet, m₁ = 0.1 kg

initial speed of bullet, u₁ = 100 m/s

mass of block, m₂ = 3 kg

initial speed of block, u₂ = 0

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Applying the principle of conservation linear momentum, for inelastic collision;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

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v = 10/3.1

v = 3.226 m/s

Part (B)

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Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²

Ki =  ¹/₂(0.1 x 100²) +  ¹/₂(3 x 0²)

Ki = 500 + 0

Ki = 500 J

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Kf = ¹/₂v²(m₁ + m₂)

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Kf = ¹/₂ x 3.226²(3.1)

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