The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground. It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.
It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.
More and more PE turns into KE as the ball falls, all the way down.
When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.
Answer:
3.08 Nm
Explanation:
N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree
The angle made with the normal of the coil, theta = 90 - 30 = 60 degree
Torque = N I A B Sin Theta
Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60
Torque = 3.08 Nm
Answer:
that best describes the process is C
Explanation:
This problem is a calorimeter process where the heat given off by one body is equal to the heat absorbed by the other.
Heat absorbed by the smallest container
Q_c = m ce (
-T₀)
Heat released by the largest container is
Q_a = M ce (T_{i}-T_{f})
how
Q_c = Q_a
m (T_{f}-T₀) = M (T_{i} - T_{f})
Therefore, we see that the smaller container has less thermal energy and when placed in contact with the larger one, it absorbs part of the heat from it until the thermal energy of the two containers is the same.
Of the final statements, the one that best describes the process is C
since it talks about the thermal energy and the heat that is transferred in the process
From the law of conservation of momentum
m1u1+ m2u2= m1v1+ m2v2
110*8+ 110*-10= 110*-10 + 110* v2
v2= 8 m/sec