Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
Substituting,
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
Where here is the velocity after the collision.
Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
Answer:
Explanation:
Molar mass of helium =
for helium is 1.67
The velocity of helium in sound at any temperature is given by
R is a constant 8.314
Answer:
His resulting velocity will be 0.187 m/s backwards.
Explanation:
Given:
Mass of the man is,
Mass of the ball is,
Initial velocity of the man is,
Initial velocity of the ball is,
Final velocity of the ball is,
Final velocity of the man is,
In order to solve this problem, we apply law of conservation of momentum.
It states that sum of initial momentum is equal to the sum of final momentum.
Momentum is the product of mass and velocity.
Initial momentum = Initial momentum of man and ball
Initial momentum =
Final momentum = Final momentum of man and ball
Final momentum =
Now, initial momentum = final momentum
The negative sign implies backward motion of the man.
Therefore, his resulting velocity is 0.187 m/s backwards.