If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.
Given,
Tension force in the rope is (T) = 160 N
Displacement of the skier (S) = 270 m
The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.
Therefore, work done by the rope on the skier is,

⇒
Hence work done by the rope is - 43200 J.
Learn more about force problems on
brainly.com/question/26850893
#SPJ4
First change km/ s into m/s, then use the formula
Lambda = velocity/ frequency
Recall the equation for magnetic force:
F = qv x B *x is cross product, not separate variable!
If the magnetic field points towards N and you throw E, then the magnetic force would point up, or out of the page. Use the right-hand rule. You point your finger towards the direction of the object, and curl your finger to the magnetic field. Your thumb is the direction of the magnetic force.
Hope this helps!
Answer:
h'=0.25m/s
Explanation:
In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).
So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of
. As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.
If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

When solving for r, we get:

so we can substitute this into our volume of a cone formula:

which simplifies to:


So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

Which simplifies to:

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)
So we get:

Now we can substitute the provided values into our equation. So we get:

so:
