You are taking the elevator to 10th floor. Are you in motion.
1 answer:
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The angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.
<h3></h3><h3>What is angular speed ?</h3>The rate of change of angular displacement is defined as angular speed. It is stated as follows:
ω = θ t
Where,
θ is the angle of rotation,
t is the time
ω is the angular velocity
The torque is found as;l
If the force is acting on the rod from the three point is the same, the value of the torque is depends upon the radius or the perpendicular distance.
The perpendicular distance of the right force is grater. Hence, the force acting on the right side is more, and the rod will rotate clockwise.
Both the forces are acting downwards. Thus, the resultant force is the less due to which the speed is increasing.
Hence, the angular speed is decreasing and direction of rotation clockwise of the rod immediately after time t.
To learn more about the angular speed, refer to the link;
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Complete question:
A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.
(a) Find the available kinetic energy of the bullet. J
(b) Find the heat required to melt the bullet. J
Answer:
Part (a) the available kinetic energy of the bullet is 323.32 J
Part (b) the heat required to melt the bullet is 216.17 J
Explanation:
Given;
mass of the bullet = 3.53 g = 0.00353 kg
velocity of the bullet = 428 m/s
initial temperature of the bullet = 40.0°C
final temperature of the bullet = 327°C
specific heat capacity, c= 128 J/(kg · °C)
latent heat of fusion, Hf = 24.5 kJ/kg
Part (a) the available kinetic energy of the bullet. J
KE = ¹/₂ × mv²
KE = ¹/₂ × 0.00353 × 428²
= 323.32 J
Part (b) the heat required to melt the bullet. J
This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.
Quantity of heat required to raise the temperature of the bullet:
Q = mcΔT
= 0.00353 × 128 × (327-40)
= 0.00353 × 128 × 287
= 129.68 J
Quantity of heat required to melt the bullet:
Q = 0.00353 × 24500
= 86.49 J
TOTAL energy required to melt the bullet = 129.68 J + 86.49 J
= 216.17 J