A power plant to represent the need for and use for energy; label it energy production.
1 answer:
You might be interested in
Answer:
the force P required for impending motion is 132.3 N
the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
Explanation:
Given that:
mass of the cabinet m = 45 kg
coefficient of static friction μ = 0.30
A free flow body diagram illustrating what the question represents is attached in the file below;
The given condition from the question let us realize that ; the casters are locked to prevent the tires from rotating.
Thus; considering the forces along the vertical axis ; we have :
The upward force and the downward force is :
where;
are the normal contact force at center point A and B respectively .
------- equation (1)
Considering the forces on the horizontal axis:
where ;
are the static friction at center point A and B respectively.
which can be written also as:
replacing our value from equation (1)
P = 0.30 ( 441)
P = 132.3 N
Thus; the force P required for impending motion is 132.3 N
b) Since the horizontal distance between the casters A and B is 480 mm; Then half the distance = 480 mm/2 = 240 mm = 0.24 cm
the largest value of "h" allowed for the cabinet is not to tip over is calculated by determining the limiting condition of the unbalanced torque whose effect is canceled by the normal reaction at and it is shifted to
:
Then:
h = 0.8 m
Thus; the largest value of "h" allowed if the cabinet is not to tip over is 0.8 m
<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased.
In fact, part of the energy has been dissipated due to the air resistance.
The mechanical energy of the ball as it starts the motion is:
</span>
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>
In fact, part of the energy has been dissipated due to the air resistance.
The mechanical energy of the ball as it starts the motion is:
</span>
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>
Answer:
Acceleration, in m/s, of such a rock fragment =
Explanation:
According to Newton's Third Equation of motion
Where:
is the final velocity
is the initial velocity
a is the acceleration
s is the distance
In our case:
So Equation will become:
Acceleration, in m/s, of such a rock fragment =