How does a steam engine do work?

What is the relation between the gravitational force acting between any two bodies object with their masses and the distance bet

LenaWriter [7]

Answer:

The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. ... increases, the force of gravity decreases. If the distance is doubled, the force of gravity is one-fourth as strong as before.

7 0

3 years ago

A metal rod of length (L) moves with velocity (v), perpendicular to its length, in a magnetic field B, which is perpendicular to

Alborosie

Explanation:

If a metal rod of length L moves with velocity v is moving perpendicular to its length, in a magnetic field B, the induced emf is given by :

$\epsilon=Blv$

The electric field in the conductor is given by :

$E=\dfrac{\epsilon}{l}\\\\E=\dfrac{Blv}{l}\\\\E=Bv$

It is clear that the electric field is independent of the length of the rod. If the length of the rod is doubled, the electric field in the rod remains the same.

6 0

3 years ago

A freight car moves along a friction less level railroad track at constant speed. The car is open on top. A large load of coal i

maxonik [38]

Answer:

The velocity of the freight car decreases.

Explanation:

This question is answered by the conservation of momentum principle.

When the freight car is moving at a certain speed, it has a constant momentum.

We will call this M1.

The equation for M1 will be:

M1 = Mass * Speed

Now when the coal is dumped into the freight car, the Mass increases.

Since conservation of momentum states that the momentum will remain the same. We have:

M1 = (Mass of freight + Mass of coal) * Speed

Since M1 is constant, if the mass increases, the speed had to decrease to keep the equation true.

8 0

3 years ago

A 975-kg sports car (including driver) crosses the rounded top of a hill at determine (a) the normal force exerted by the road o

iVinArrow [24]

There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill: $v=15 m/s$
- radius of the hill: $r=100 m$

Solution:

(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car $W=mg$ (downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, $m \frac{v^2}{r}$ , so we can write:
$mg-N=m \frac{v^2}{r}$ (1)
By rearranging the equation and substituting the numbers, we find N:
$N=mg-m \frac{v^2}{r}=(975 kg)(9.81 m/s^2)-(975 kg) \frac{(15 m/s)^2}{100 m}=7371 N$

(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
$N=mg-m \frac{v^2}{r}=(62 kg)(9.81 m/s^2)-(62 kg) \frac{(15 m/s)^2}{100 m}=469 N$

(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
$mg=m \frac{v^2}{r}$
from which we find
$v= \sqrt{gr}= \sqrt{(9.81 m/s^2)(100 m)}=31.3 m/s$

7 0

4 years ago

If a proton is released at the equator and falls toward earth under the influence of gravity, the magnetic force on the proton w

pantera1 [17]

Positive positive side naturally because protons always move to positive side and electron always move to negative side

3 0

3 years ago