Answer:
15009
Explanation:
PE = mgh
PE = 61.2(9.81)(10 * 2.50)
PE = 15009.3
The force needed to accelerate an elevator upward at a rate of
is 2000 N or 2 kN.
<u>Explanation:
</u>
As per Newton's second law of motion, an object's acceleration is directly proportional to the external unbalanced force acting on it and inversely proportional to the mass of the object.
As the object given here is an elevator with mass 1000 kg and the acceleration is given as
, the force needed to accelerate it can be obtained by taking the product of mass and acceleration.


So 2000 N or 2 kN amount of force is needed to accelerate the elevator upward at a rate of
.
Answer:true,because velocity is directly proportional to speed or velocity
Explanation:
Velocity = frequency x wavelength
The velocity or speed varies directly with the frequency, so as the frequency is increased, the velocity or speed is also increased
Answer:
f1= -350cm or -3.5m
f2= 22.1cm or 0.221m
Explanation:
A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.
Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.
Answer:
The ball would hit the floor approximately
after leaving the table.
The ball would travel approximately
horizontally after leaving the table.
(Assumption:
.)
Explanation:
Let
denote the change to the height of the ball. Let
denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let
denote the initial vertical velocity of this ball.
If the air resistance on this ball is indeed negligible:
.
The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was
.
The height of the table was
. Therefore, after hitting the floor, the ball would be
below where it was before leaving the table. Hence,
.
The equation becomes:
.
Solve for
:
.
In other words, it would take approximately
for the ball to hit the floor after leaving the table.
Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at
) until the ball hits the floor.
The ball was in the air for approximately
and would have travelled approximately
horizontally during the flight.