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N76 [4]
2 years ago
7

A rocket fires two engines simultaneously. One produces a thrust of 725Ndirectly forward while the other gives a 513N thrust at

32.4° above the forward direction. Find the magnitude and direction (relative to the forward direction) of the resultant force which these engines exert on the rocket.
Physics
1 answer:
faust18 [17]2 years ago
8 0

The magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

<h3>What is the resultant force the two engines exert on the rocket?</h3>

The resultant force on the rocket is calculated thus:

The 513N thrust is resolved into vertical and horizontal components;

Horizontal component: 513N cos(32.4°) = 433.14 N

Vertical component: 513N sin(32.4°) = 274.88 N

Total forward force on the rocket = 725 N + 433.14 N = 1,158.14 N

Total force at right angles:

0 + 274.88 N = 274.88 N

The resultant force (F) is then given as follows:

F² = a² + b²

F² = (1158.14 N)² + (274.88 N)²

F = √1,416,847.27

F = 1,190.3

To find the direction:

tan X 274.88 N / 1,158.14 N

X = tan⁻¹ 0.237346089419241

X = 13.35°

Therefore, the magnitude of the resultant force, F = 1,190.3 acting at a direction X = 13.35°.

In conclusion, the resultant force is obtained by resolving the forces into vertical and horizontal components.

Learn more about resultant force at: brainly.com/question/17434363

#SPJ1

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Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
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Answer:

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Explanation:

We are given that

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We have to find the exit temperature.

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C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

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1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

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T_2=500+1.598-115.4

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