Event of movement pearl harbor speech

1 answer:

7 0

On December 7th, 1941, Pearl Harbor was attacked by Japanese forces. The next day, Franklin D. Roosevelt addressed the United States Congress with his memorable “a date which will live in infamy” speech.This speech had two purposes:To rally the American people to support the war effort.

You might be interested in

Which material is colder in a room where all the objects are in for a long time?

Ahat [919]

Wooden desk it doesn’t conduct or absorb heat as much as metal or plastic

3 0

3 years ago

How can gravitational tugs from orbiting planets affect the motion of a star?.

gayaneshka [121]

Answer:

Gravitational tugs from orbiting planets don't affect the motion of a star. The star, being much larger than the planet, has a much smaller orbit. But it does move slightly. Explain how alien astronomers could deduce the existence of planets in our solar system by observing the Sun's motion.

3 0

2 years ago

Oxygen is a reactant in a combination reaction, which always produces energy in the form of heat and light. A. True B. False

harkovskaia [24]

The answer is B.False

8 0

3 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:

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