Answer: Find the answer in the explanation
Explanation: Given the Roman numeral and the representation
I. part of a coal-fired power plant
II. part of a nuclear power plant
III. part of a coal-fired power plant and part of a nuclear power plant
a.) Boiler : I
b.) Combustion chamber: I
c.) Condenser: I
d.) Control rod: II
e.) Generator: III
f.) Turbine: III
Toward the end processes part of both coal fire and nuclear power, they both make use of turbine and generator to generate electricity.
a) 2.75 s
The vertical position of the ball at time t is given by the equation

where
h = 4 m is the initial height of the ball
u = 12 m/s is the initial velocity of the ball (upward)
g = 9.8 m/s^2 is the acceleration of gravity (downward)
We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:

This is a second-order equation. By solving it for t, we find:
t = -0.30 s
t = 2.75 s
The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.
b) -15.0 m/s (downward)
The final velocity of the ball can be calculated by using the equation:

where
u = 12 m/s is the initial (upward) velocity
g = 9.8 m/s^2 is the acceleration of gravity (downward)
t is the time
By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:

And the negative sign means the direction is downward.
Answer:
B
Explanation:
From Newton's law of motion, we have:
V^2 = U^2 + 2gH
Where V and U are final and initial velocity respectively.
H is the height.
For the object to have a sustain a maximum height it means the final velocity of the object is zero.
By computing the height of the object sustain by A, we have:
0^2 = 2^2 -2×10×H
0= 4 -20H
4 = 20H;
H= 0.2m
For object B we have;
0^2 = 1^2 -2×10×H
0 = 1 -20H
H = 1/20= 0.05m
From computing the height sustain by both objects, we see object B is projected at a shorter height into atmosphere than A.
Hence object B will return to the ground first.
A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>