It would be oraganic matter I think.
Final velocity = initial velocity + acceleration * time
v = u + at
v = 3.28 + 2.32 * 2.08
v = 3.28 + 4.83
<u>v = 8.11 m/s</u>
Answer:
Explanation:
It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.
Let in equilibrium , tension in rope be T
For balancing Joe
T = M g
For balancing Simon
friction + T = mgsinθ
μmgcosθ+T = mgsinθ
μmgcosθ+Mg = mgsinθ
M = (msinθ - μmcosθ)
M = m(sinθ - μcosθ)
Answer:
Explanation:
Mass of first cart M1=2.4kg
Velocity of first cart U1=4.1m/s
Mass of second cart M2=1.7kg
Second cart is initially at rest U2=0
After an instant, the velocity of the second cart is U2=-2.8m/s
Now after collision the two cart move together with the same velocity I.e inelastic collision
Using conservation of momentum
Momentum before collision, = momentum after collision
M1U1 + M2U2 = (M1+M2)V
2.4×4.1 + 1.7× -2.8 =(2.4+1.7)V
9.84 - 4.76 = 4.1V
5.08=4.1V
V=5.08/4.1
V=1.24m/s
The momentum of the two cart at that instant is
M1U1+M2U2
2.4×4.1 + 1.7× -2.8
9.84 - 4.76
5.08kgm/s
So the momentum at the instant the velocity is 4.1m/s for cart 1 and -2.8m/s for cart 2 is 5.08kgm/s