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fiasKO [112]
3 years ago
14

A space probe is traveling in outer space with a momentum that has a magnitude of 3.87 x 107 kg·m/s. A retrorocket is fired to s

low down the probe. It applies a force to the probe that has a magnitude of 1.41 x 106 N and a direction opposite to the probe's motion. It fires for a period of 9.55 s. Determine the momentum of the probe after the retrorocket ceases to fire.
Physics
1 answer:
Virty [35]3 years ago
6 0
Force = change in momentum / time, Force in opposite direction so negative
-1.41 x 10⁶ = (p₂ - 3.87 x 10⁷) / 9.55
p₂ = 2.52 x 10⁷ Ns
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Answer:

\theta = n\pi/2, {\rm where~n~is~an~integer.}

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We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k

Similarly,

\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)

So,

\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)

At time t = 0, this equation becomes

0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}

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2 years ago
How do I solve this​
Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

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W = 1,602 × 10^(-19) * 500

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Replacing we have,

\eta = 1-\frac{T_L}{T_H}

\eta = 1-\frac{293}{713}

\eta = 0.589

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