Answer:
D
Explanation:
A) is not correct, because the gravitation potential energy will depend on the height the block is located at. It will be calculated with the formula:
U=mgh.
If we take the ground as a zero height reference, then on point 2 the potential energy will be:


While on point 3, the potential energy will be greater.


B) is not the right answer because the kinetic energy will vary with the height the block is located at in the fact that the energy is conserved (this is if we don't take friction into account or air resistance) so in this case:

We already know the potential energy at point 2. We can calculate the kinetic energy at point 3 like this:



So the kinetic energy at point 2 is given by the equation:

so:


As you may see the kinetic energy at point 2 is greater than the kinetic energy at point 3.
C) Is not correct because according to the first law of thermodinamics, energy is not lost, only transformed. So, since we are not taking into account friction or any other kind of loss, then we can say that the amount of mechanical energy at point 1 is exactly the same as the mechanical energy at point 3.
D) Because of what we talked about on part C, this will be the true situation, because the mechanical energy of the block will be the same no matter theh point you measure it at.
Carson's Rule says:
FM occupied bandwidth =
(2) · (Peak deviation + Highest modulating frequency)
so
FM bw = (2) · (60 kHz + 15 kHz)
FM bw = (2) · (75 kHz)
FM bw = 150 kHz
(I used to eat this stuff for lunch, but it's been almost 40 years. Thanks for taking me back. Those were the good old days.)
Answer
given,
V = 2 L
the left is an ideal gas at P = 100 k Pa and T = 500 K
mass is constant


Pressure is same because it's not changing due to process






m = 1.39 x 10⁻³ Kg


Answer:
Alcohol and drugs are a no no. you should put the answers in so people can answer correctly.
Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;

where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive

Therefore, the normal force the seat exerted on the driver is 125 N.