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3 years ago

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A 540 kg satellite moves through deep space with a speed of 27 m/s. A booster rocket on the satellite fires for 1.4 s, giving a

force to the satellite. The speed of the rocket increases to 31 m/s. Calculate the force given by the booster rocket to the satellite. (Assume that the mass of the satellite remains constant.)
A)1543 N
B)2160 N
C)11,957 N
D)3024 N

2 answers:

prohojiy [21]3 years ago

6 0

Answer: a

Explanation: because the answer is 1.4444444 and that's the closest

Tema [17]3 years ago

5 0

Answer: A. Hope this helped you

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Part a)

$T = 342.5 \hat i + 675\hat j$

Part b)

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Part c)

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Part d)

$\theta = 333 degree$

Part e)

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Part f)

$\theta = 333 degree$

Explanation:

Part a)

Magnitude of tension force is given as

$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

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Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

$F_{net} = 342.5 \hat i + 675\hat j - 849 \hat j$

$F_{net} = 342.5\hat i - 174\hat j$

Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

$F = \sqrt{342.5^2 + 174^2}$

$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

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$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

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Learn more about dynamic pressure here:

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