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CaHeK987 [17]
3 years ago
11

An astronaut working with many tools some distance away from a spacecraft is stranded when the "maneuvering unit" malfunctions.

How can the astronaut return to the spacecraft by sacrificing some of the tools? (Note: the maneuvering unit is connected to the astronaut's spacesuit and is not removable)
Physics
1 answer:
N76 [4]3 years ago
6 0

Answer:

He can return to the spacecraft by sacrificing some of the tools employing the principle of conservation of momentum.

Explanation:

By carefully evaluating his direction back to the ship, the astronaut can throw some of his tools in the opposite direction to that. On throwing those tools of a certain mass, they travel at a certain velocity giving him velocity in the form of recoil in the opposite direction of the velocity of the tools. This is same as a gun and bullet recoil momentum conservation. It is also the principle on which the operational principles of their maneuvering unit is designed.

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A 2020 kg car traveling at 14.2 m/s collides with a 2940 kg car that is initally at rest at a stoplight. The cars stick together
velikii [3]

Answer:

The answer is 3,064xe^{-4}

Explanation:

When the collision happens, the momentum of the first car is applied to the both of them.

So we can calculate the force that acts on both cars as:

  • The momentum of the first car is P = 2020 kg x 14.2 m/s = 28,684 kg.m/s
  • The acceleration of both cars after the crash is going to be a = P / mtotal which will give us a = 28,684 / (2020+2940) = 5.78 m/s
  • Since the second car was initially not moving, the final acceleration was calculated with the momentum of the first car.

Now we can find the force that acts on both of them by using the formula F = m.a which will give us the result as:

  • F = (2020+2940) x 5.78 = 28,684

The friction force acts in the opposite direction and if they stop after moving 2.12 meters;

  • Friction force is Ff = μ x N where μ is the friction coefficient and the N is the normal force which is (2020+2940) x 10 if we take gravitational force as 10, equals to 49,600.
  • F - Ffriction = m x V
  • 28,684 - μ x 49,600 = 4960 x 5.78
  • μ = 3,064xe^{-4}

4 0
3 years ago
Relative to a stationary observer, a moving clock Group of answer choices can do any of the above. It depends on the relative en
Schach [20]

Answer:

always runs slower than normal.

Explanation:

The basic concept of theory of relativity was given famous scientist, Albert Einstein. The relativity theory provides the theory of space and time, which are the two aspects of spacetime.

According to the theory of relativity, the laws of physics are same for all the non-accelerating observers.

In the context, according to the theory of relativity, a moving clock relative tot a stationary observer always runs slower than the normal time.

7 0
2 years ago
A cube and a square pyramid were joined to form the composite solid. A cube with side lengths of 12 inches. A square pyramid wit
Vesna [10]

Answer:

i think it is the first option.

6 0
3 years ago
Read 2 more answers
Which of the following affects the rate constant of a reaction?
Art [367]
A the entropy of the reaction I think ion know if that’s correct
3 0
1 year ago
Two balloons are charged with an identical quantity and type of charge: -4 nc They are held apart at a separation distance of 70
blagie [28]

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

F=k\frac{q_{1} q_{2} }{r^{2}}

where k is a proportionality constant known as the Coulomb's law constant. Its value is 9 \times 10^{9} Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = 4 \times 10^{-9} C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N

Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN

8 0
3 years ago
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