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anzhelika [568]
2 years ago
6

A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k

g of exhaust at a speed of 230 m/s relative to the space probe. What is the final velocity of the probe?
Physics
1 answer:
Reika [66]2 years ago
7 0

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

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Answer:

F=-100N; a=1.3m/s^2

Explanation:

Force is being made by student, so wall counteracts that force by not moving so it is equally opposite.

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3 years ago
wo kids are on a seesaw. The one on the left has a mass of 75 kg and is sitting 1.5 m from the pivot point. The one on the right
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Answer:

4.5 Nm (Anticlockwise)

Explanation:

Let the 75 kg kid is sitting at the left end and the 60 kg kid is sitting on the right end.

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clockwise Torque = 60 x 1.8 = 108 Nm

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2 years ago
Analyze the data to identify the mathematical relationship between the
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2 years ago
The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

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q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

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Hope this helps! :)
8 0
3 years ago
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