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1 year ago

An astronaut leaves a spaceship that is moving through free space. Will the spaceship move off and leave the astronaut behind?

1 answer:

9 0

When an astronaut leaves a spaceship that is moving through free space, the spaceship does not move off and leave the astronaut behind.

This is because when the astronaut is inside the spaceship, which is moving in free space, the astronaut also has the same velocity as the spaceship. Hence, when he leaves the spaceship, there is no relative velocity between the spaceship and the astronaut. This is the reason why the spaceship will not move off and leave the astronaut behind.

A spaceship or a spacecraft is an especially designed vehicle that is used to travel to outer space. It can be used for space exploration, transportation of humans and cargo, and meteorological purposes.

An astronaut is a person who has undergone specific training to travel to space. The duties of an astronaut aboard a spaceship can vary. Usually, a pilot and a commander work together to lead the mission. Other positions could be science pilot, mission expert, payload commander, and flight engineer.

Refer to more about astronaut here

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Answer:

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3 years ago

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A laser pulse of duration 25 ms has a total energy of 1.4 J. The wavelength of this radiation is

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Answer:

n = 4 x 10¹⁸ photons

Explanation:

First, we will calculate the energy of one photon in the radiation:

$E = \frac{hc}{\lambda}\\\\$

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$E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{5.67\ x\ 10^{-7}\ m}$

E = 3.505 x 10⁻¹⁹ J

Now, the number of photons to make up the total energy can be calculated as follows:

$Total\ Energy = nE\\1.4\ J = n(3.505\ x\ 10^{-19}\ J)\\n = \frac{1.4\ J}{3.505\ x\ 10^{-19}\ J}\\$

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3 years ago

A cannon, elevated at 40∘ is fired at a wall 300 m away on level ground, as shown in the figure below. The initial speed of the

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Vo = 89 m/s
angle: 40°

=> Vox = Vo * cos 40° = 89 * cos 40°

=> Voy = Vo. sin 40° = 89 * sin 40°

x-movement: uniform => x =Vox * t = 89*cos(40)*t

x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s

y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2

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A 1.25 kg block is attached to a spring with spring constant 17.0 N/m . While the block is sitting at rest, a student hits it wi

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Answer:

The amplitude of the subsequent oscillations is 13.3 cm

Explanation:

Given;

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speed of the block, v = 49 cm/s = 0.49 m/s

To determine the amplitude of the oscillation.

Apply the principle of conservation of energy;

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$K.E_{max} = U_{max}\\\\\frac{1}{2} mv^2 = \frac{1}{2} kA^2\\\\where;\\\\A \ is \ the \ maximum \ displacement = amplitude \\\\mv^2 = kA^2\\\\A^2 = \frac{mv^2}{k} \\\\A = \sqrt{\frac{mv^2}{k}} \\\\A = \sqrt{\frac{1.25\ \times \ 0.49^2}{17}} \\\\A = 0.133 \ m\\\\A = 13.3 \ cm$

Therefore, the amplitude of the subsequent oscillations is 13.3 cm

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3 years ago