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3 years ago

13

The motor of a four wheeler traveling along a muddy trail generates an average power of 7.48 104 W when moving at a constant spe

ed of 13 m/s. When pulling a log along the trail at the same speed, the engine must generate an average power of 8.50 104 W. What is the tension in the tow rope pulling the log?

2 answers:

Radda [10]3 years ago

7 0

Answer:

The tension in the tow rope pulling the log is 784.61 N

Explanation:

Given:

v = speed = 13 m/s

Pf = final power when the log is pulled = 8.5x10⁴W

Pi = average power = 7.48x10⁴W

The force required to move a car is equal to:

$F_{i} =\frac{P_{i} }{v} =\frac{7.48x10^{4} }{13} =5753.85 N$

$F_{f} =\frac{P_{f} }{v} =\frac{8.5x10^{4} }{13} =6538.46 N$

T = Ff - Fi = 6538.46 - 5753.85 = 784.61 N

DiKsa [7]3 years ago

7 0

Answer: 784.6 N

Explanation:

Given

Power of the motor, P1 = 7.48*10^4 W

Speed of the motor, v = 13 m/s

Power of the motor, P2 = 8.5*10^4 W

It is known that

Power = J/t, where

J = work, in joules

t = time, in seconds. Also,

Work = F * d, where

F = force

d = distance, thus on substituting, we have,

Power = Fd/t

Also, we know that

Velocity = d/t, where

d = distance,

t = time, thus, we could say that

Power = Force * Velocity

P = FV

To find the tension in the log, we use

F = P/V, where

P = P2 - P1

P = 8.5*10^4 - 7.48*10^4

P = 1.02*10^4 = 10200W

Force, F = 10200 / 13

F = 784.6 N

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Answer

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Answer:

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Explanation:

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Can anyone help me? (physics)

Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

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Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

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Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

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Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

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U = 13.23 / Sine θ

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U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

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