Can someone please help me to compare and contrast Gravitational force and Electromagnetic force?

Determine for which class of lever the output force is always greater than the input force. for which class is the output force

umka2103 [35]

The 3rd class lever is the <span>output force always less than the input force, becuase its mechanical advantage is always less than 1. this also due that in a 3rd class lever the effort arm is shorter than the load arm, that is why the output is lower than the input force. but 3rd class lever is a speed multiplier lever</span>

5 0

4 years ago

Read 2 more answers

A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a

lina2011 [118]

Answer:

hinit = 17.5 m

Explanation:

Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

$E_{o} = U_{o} = m*g*h_{init} (1)$

When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
We can express this final energy as follows:

$E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)$

In order to find hinit, we need to make (1) equal to (2), and solve for it.
In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
So, the centripetal force must be equal to the weight.
The centripetal force can be expressed as follows:

$F_{c} = m*\frac{v_{f} ^{2}}{R} (3)$

We have just said that (3) must be equal to the weight:

$F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)$

Simplifying, and rearranging, we can solve for vf², as follows:

$v_{f}^{2} = R*g (5)$

Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

$h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)$

7 0

3 years ago

A dynamite blast propels a heavy rock straight up with a launch velocity of 160ft/sec (about 109 mph). Write a position for the

Flura [38]

A) Using:
2as = v² - u², where v will be 0 at max height
s = -(160)² / 2 x -32.174
s = 397.8 ft

b) Using:
s = ut + 1/2 at²
256 = 160t - 16.1t²
solving for t,
t = 2.0, t = 7.9
Now, v = u + at, for both times:
v(2) = 160 - 32.174(2)
v(2) = 95.7 ft/sec on the way up

v(7.9) = 160 - 32.174(7.9)
v(7.9) = -94.3 ft/sec; 94.3 ft/sec on the way down

c) -32.174 ft/s², which is the acceleration due to gravity.

d) s = 0
0 = 160t - 1/2 x 32.174t²
t = 9.94 seconds

3 0

4 years ago

What can be caused by nuclear explosions and can easily penetrate into the human body causing severe damage?

dedylja [7]

Answer:

Nuclear explosions produce air-blast effects similar to those produced by conventional explosives. The shock wave can directly injure humans by rupturing eardrums or lungs or by hurling people at high speed, but most casualties occur because of collapsing structures and flying debris.

8 0

3 years ago

Read 2 more answers

The force required to maintain an object at a constant velocity in free space is equal to

alex41 [277]

Answer:

The force required to maintain an object at a constant velocity in free space is equal to Zero.

7 0

3 years ago