Its during REM sleep. hope this helped
Answer:

Explanation:
Mass of the ship (m) = 6.9 × 10⁷ kg
Speed of the ship (v) = 33 km/h
First, let us convert the speed from km/h to m/s using the conversion factor.
We know that, 1 km/h = 5/18 m/s
So, 33 km/h = 
Now, we know, the momentum of an object only depends on its mass and speed. Momentum is independent of the length of the object.
So, here, length of the ship doesn't play any role in the determination of the momentum.
Magnitude of momentum of the ship = Mass × Speed
= 
= 
Therefore, the magnitude of ship's momentum is
.
Answer:
The maximum static frictional force is 40N.
Explanation:
When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).
This coefficient defines the maximum static friction force that we can have.
So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.
In this case, we know that we apply a force of 40N and the object just starts to move.
Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.
Hello!
Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.
Data:
Hooke represented mathematically his theory with the equation:
F = K * Δx
On what:
F (elastic force) = 2 N
K (elastic constant) = 4 N/cm
Δx (deformation or elongation of the elastic medium or distance from a spring) = ?
Solving:




simplify by 2


Answer:
B.) 1/2 cm
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I Hope this helps, greetings ... Dexteright02! =)