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nikitadnepr [17]
2 years ago
14

An astronaut with mass 84kg is at rest in space, before firing her water pack to move toward the shuttle. If the amount of water

shot out is
2kg and it is fired at a speed of 10m/s, what will the speed of the astronaut be?
Physics
1 answer:
Travka [436]2 years ago
4 0

Answer:

The astronaut is moving at a speed of 0.238 m/s in a direction opposite the direction of the water shot out.

Explanation:

We are given;

Mass of astronaut; m1 = 84 kg

Mass of water shoot out; m2 = 2 kg

Initial speed of astronaut; u1 = 0 m/s

Initial speed of water shoot out; u2 = 0 m/s

Final speed of shoot out; v2 = 10 m/s

From law of conservation of momentum, we can say that;

Initial momentum = final momentum

Thus;

m1•u1 + m2•u2 = m1•v1 + m2•v2

Where v1 is the final speed of the astronaut

Plugging in the relevant values, we get;

(84 × 0) + (2 × 0) = (84 × v1) + (2 × 10)

0 = 84v1 + 20

-20 = 84v1

v1 = -20/84

v1 = -0.238 m/s

The negative sign indicates that the astronaut is moving 0.238 m/s in a direction opposite the direction of the water shot out.

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A mass of 240 grams oscillates on a horizontal frictionless surface at a frequency of 2.5 Hz and with amplitude of 4.5 cm.
mr_godi [17]

Answer:

(a) The spring constant is 59.23 N/m

(b) The total energy involved in the motion is 0.06 J

Explanation:

Given;

mass, m = 240 g = 0.24 kg

frequency, f = 2.5 Hz

amplitude of the oscillation, A = 4.5 cm = 0.045 m

The  angular speed is calculated as;

ω = 2πf

ω = 2 x π x 2.5

ω = 15.71 rad/s

(a) The spring constant is calculated as;

\omega = \sqrt{\frac{k}{m} } \\\\\omega ^2 = \frac{k}{m} \\\\k = m\omega ^2\\\\where;\\\\k \ is \ the \ spring \ constant\\\\k = (0.24) \times (15.71)^2\\\\k = 59.23 \ N/m

(b) The total energy involved in the motion;

E = ¹/₂kA²

E = (0.5) x (59.23) x (0.045)²

E = 0.06 J

5 0
3 years ago
Please help me answer this​
ivann1987 [24]

Answer:

A is the correct answer

Hope it helps

4 0
2 years ago
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