Answer:
I'm pretty sure it's 3.
Explanation:
Because if you look at your options the only that would be relevant to tick marks would be either 4 or 3. And it said in the question that we're looking for the one for the dependent variable. And the dependent variable is on the Y- Axis and the 3 is the tick marks for the y-axis. So your answer is 3.
Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J
Answer:
Option (4)
Explanation:
There are two types of collision.
Perfectly elastic collision: the collision in which the momentum and kinetic energy is conserved. There is no loss of energy in other forms of energy.
Perfectly plastic collision: The collision in which the momentum is conserved and kinetic energy is not conserved. The two bodies stick after the collision.
Here, the bullet hits the block and then embedded in the block, it is the example of plastic collision.
