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bezimeni [28]
3 years ago
6

A space walking astronaut has become detached from her spaceship.shes floating in space while holding a wrench she is thinking a

bout how to get back to the spaceship.which she can see 50m from her current location what can she do to get back to the spaceship? Explain your answer using Newton’s 3 laws of motion
Physics
1 answer:
victus00 [196]3 years ago
4 0

Answer:

Newton's third law of motion states that for every action, there is equal and opposite reaction.

While space walking, when the astronaut gets detached from the space ship, she floats in space holding a wrench. In order to get back to the spaceship, she should throw the wrench in the opposite direction of the spaceship. This action would  cause a reaction on her own body and she would be pushed away from the wrench and towards the spaceship. Thus, she can return back to the spaceship in this way.

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If a 100N weight sits in an express elevator what will be the weight when the elevator is traveling at a constant speed
yarga [219]
This is a "trick" question.

If the elevator is traveling at constant speed, it means it is at rest. This means anything inside the elevator traveling at constant speed, weights the same as in an elevator not moving -also at rest-.

So the 100N weight's weight doesn't change in an elevator traveling at constant speed.
4 0
3 years ago
Which groups of organisms became extinct during the Paleozoic Era
guapka [62]
Dinosaurs but I need the whole groups yo tell you ;)
4 0
3 years ago
If the force of attraction (gravity) on the moon is 1/6 that of the force on Earth, what would
Aloiza [94]

I believe that you would weigh around 68 or 69 N, or 7 kilograms.

4 0
2 years ago
A 55-kg woman is wearing high heels. If each heel has a circular cross-section 6.0 mm in diameter and she puts all her weight on
bogdanovich [222]

Answer:

The pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

Explanation:

Given;

mass of the woman, m = 55 kg

diameter of the circular heel, d = 6.0 mm

radius of the heel, r = 3.0 mm = 0.003 m

Cross-sectional area of the heel is given by;

A = πr²

A = π(0.003)²

A = 2.8278 x 10⁻⁵ m²

The weight of the woman is given by;

W = mg

W = 55 x 9.8

W = 539 N

The pressure exerted by the woman on the floor is given by;

P = F / A

P = W / A

P = 539 / (2.8278 x 10⁻⁵ )

P = 1.9061 x 10⁷ N/m²

Therefore, the pressure exerted by the woman on the floor is 1.9061 x 10⁷ N/m²

5 0
3 years ago
Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface
FrozenT [24]

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

\tau=I\alpha     (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

\tau=Fd        (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

Fd=I\alpha\\\\I=\frac{Fd}{\alpha}

Finally, you replace the values of all parameters in the previous equation for I:

I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2

The moment of inertia of the door around the hinges is 2 kgm^2

3 0
3 years ago
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