Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s
Newton's second law states that 
(Force is mass times acceleration). Out of the given options, “a 10 kg ball thrown with a 50 Newton force”, an example have the greatest acceleration.
<u>Explanation:
</u>
To do the calculation, we already have the formula given derived from Newton’s second law of motion. To know the acceleration, we can simply modify the formula as,
For the 10kg ball threw in 50N, we have mass = 10 kg and Force = 50 N
Acceleration = 
Similarly for 1 kg ball threw in 0.5N, substituting the values, we get
,
Acceleration = 
For launching 50kg student by catapult of 100N,
acceleration = 
For accelerating 500 kg car in 1000N engine,
acceleration = 
Answer:
A= 150 J
Explanation:
Kinetic energy is the energy of an object in motion.
The formula for kinetic energy is ;
K.E = 1/2 * m *v² where m is mass and v is velocity
Work done is equal to change in kinetic energy
W= Δ K.E
Given that K.E = 150 J
Taking that the ball was stationary before it was thrown, this makes its initial kinetic energy to be 0 J so the work done will be
W= Δ K.E
W= 150 - 0
W= 150 J
One with greater mass (8kg)
We have that a blackbody radiator either constantly absorbs energy or constantly emits energy, depending on its surroundings. In this case, the energy is continuously and smoothly decreasing, thus it cannot be like B and C.
The energy loss or gain is also monotonous, it has the same direction; a radiator cannot gain energy at some point and then lose some. Hence, it does not resemble a wave either. The most appropriate model is the ramp. Energy is constantly emitted to surroundings and it decreases monotonically.
