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dalvyx [7]
3 years ago
15

A rocket, weighing 43576 N, has an engine that provides an upward force of 11918 N. It reaches a maximum speed of 713 m/s. For h

ow much time must the engine burn during the launch in order to reach this speed? Call up positive.
Physics
1 answer:
vladimir1956 [14]3 years ago
8 0
Answer: 266.0 s ≈ 4.4 min

Explanation:

1) Data

Wr = 43,576 N
F = 11,918 N
V = 713 m/s
t =?

2) Principles and formulas

Impulse and conservation of momentum

I = F.t = Δp

Δp = mΔv

3) Solution

m = Wr / g = 43,576N / 9.8 m/s^2 = 4,446.5 kg

Δp = mΔv => 4,446.5 kg * 713 m/s = 3,170,354,5 N*s

I = F.t = Δp  => t = Δp / F = 3,710,354.5 N*s / 11,918N = 266.0 s

t = 266.0 s ≈ 4.4 min
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98 m √

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P-weight blocks D and E are connected by the rope which passes through pulley B and are supported by the isorectangular prism ar
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Answer:

21.8°

Explanation:

Let's call θ the angle between BC and the horizontal.

Draw a free body diagram for each block.

There are 4 forces acting on block D:

Weight force P pulling down,

Normal force N₁ pushing perpendicular to AB,

Friction force N₁μ pushing parallel up AB,

and tension force T pushing parallel up AB.

There are 4 forces acting on block E:

Weight force P pulling down,

Normal force N₂ pushing perpendicular to BC,

Friction force N₂μ pushing parallel to BC,

and tension force T pulling parallel to BC.

Sum of forces on D in the perpendicular direction:

∑F = ma

N₁ − P sin θ = 0

N₁ = P sin θ

Sum of forces on D in the parallel direction:

∑F = ma

T + N₁μ − P cos θ = 0

T = P cos θ − N₁μ

T = P cos θ − P sin θ μ

T = P (cos θ − sin θ μ)

Sum of forces on E in the perpendicular direction:

∑F = ma

N₂ − P cos θ = 0

N₂ = P cos θ

Sum of forces on E in the parallel direction:

∑F = ma

N₂μ + P sin θ − T = 0

T = N₂μ + P sin θ

T = P cos θ μ + P sin θ

T = P (cos θ μ + sin θ)

Set equal:

P (cos θ − sin θ μ) = P (cos θ μ + sin θ)

cos θ − sin θ μ = cos θ μ + sin θ

1 − tan θ μ = μ + tan θ

1 − μ = tan θ μ + tan θ

1 − μ = tan θ (μ + 1)

tan θ = (1 − μ) / (1 + μ)

Plug in values:

tan θ = (1 − 0.4) / (1 + 0.4)

θ = 23.2°

∠BCA = 45°, so the angle of AC relative to the horizontal is 45° − 23.2° = 21.8°.

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3 years ago
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