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Arada [10]
3 years ago
7

If the first stage provides a thrust of 5.31 ​mega-newtons [MN] and the space shuttle has a mass of 5,760,000 ​pound-mass ​[lbm​

], what is the acceleration of the spacecraft in miles per hour squared ​[mi/h2​]?
Physics
1 answer:
zysi [14]3 years ago
5 0

Answer:

The acceleration of the spacecraft is 16,367.62 miles/hr²

Explanation:

Given;

thrust of the spacecraft, F = 5.31 x 10⁶ N

mass of the spacecraft, m = 5,760,000 lbm = 2612692.051 kg

The acceleration of the spacecraft in m/s² is given by Newton's second law of motion;

F = ma

a = F / m

a = \frac{5.31*10^6 \ N}{2612692.051 \ kg}\\\\a = 2.0324 \ m/s^2

The acceleration in miles per hour square is given by;

a =2.0324 \frac{m}{s^2} *\frac{0.0006214 \ mile}{m}*(\frac{3600 \ s}{1 \ hour})^2\\\\  a =2.0324 \frac{m}{s^2} *\frac{0.0006214 \ mile}{m}*\frac{12960000 \ s^2}{1 \ hour^2}\\\\a = 16,367.62 \ miles/hr^2

Therefore, the acceleration of the spacecraft is 16,367.62 miles/hr²

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Calculate the magnitude of the force exerted by each wire on a 1.20-m length of the other.
Zielflug [23.3K]

Incomplete question.The complete question is attached below as screenshot along with figure

Answer:

F=6.00*10^{-6}N

Force is repulsive

Explanation:

Given data

Current I₁=5.00A

Current I₂=2.00A

Length L=1.20 m

Radius r=0.400m

To find

Force F

Solution

As the force is repulsive because currents are in opposite direction

From repulsive force we know that:

F=\frac{u_{o}I_{1}I_{2}L}{2\pi r}

Substitute the given values

F=\frac{u_{o}(5.00A)(2.00A)(1.20m)}{2\pi (0.400m)}\\ F=6.00*10^{-6}N

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3 years ago
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