Answer:
(a) v = 15m/a
(b) No they won't feast because the rock can only rise to a height of 11.5m which is less than 15m.
Explanation:
Please see the attachment below for film solution.
The bottom of the inner curve of a hook is called
1 answer:
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Answer:
Speed of ball A after collision is 3.7 m/s
Speed of ball B after collision is 2 m/s
Direction of ball A after collision is towards positive x axis
Total momentum after collision is m×4·21 kgm/s
Total kinetic energy after collision is m×8·85 J
Explanation:
<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>Let the mass of each ball be m kg
v be the velocity of ball A along positive x axis
v be the velocity of ball A along positive y axis
u be the velocity of ball B along positive y axis
Conservation of momentum along x axis
m×3·7 = m× v
∴ v = 3.7 m/s along positive x axis
Conservation of momentum along y axis
m×2 = m×u + m× v
2 = u + v → equation 1
∴ relative velocity of approach = relative velocity of separation
-2 = v - u → equation 2
By adding both equations 1 and 2 we get
v = 0
∴ u = 2 m/s along positive y axis
Kinetic energy before collision and after collision remains constant because it is an elastic collision
Kinetic energy = (m×2² + m×3·7²)÷2
= 8·85×m J
Total momentum = m×√(2² + 3·7²)
= m× 4·21 kgm/s
Answer:
- the forces on the left hand side is 1.038 kN
- the forces on the right hand side is 1.483 kN
Explanation:
Given the data in the question, as illustrated in the image below;
Length of the scaffold = 3 m
weight of the scaffold = 395 N
Weight of Bob = 805 N and stands 1 m from the left end
weight of washing equipment = 500N and on sits 2 m from the left end
Weight of Joe = 820 N and stand 0.500 m from the right end
so the force on the left cable will be;
=
[ (805 N)( (3-1) m) + ( 395 N )(
) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]
=
[ 1610 + 592.5 + 500 + 410 ]
=
[ 3112.5 ]
= 1037.5 N
= 1.038 kN
Therefore, the forces on the left hand side is 1.038 kN
On the right hand side;
= ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N
= 2520 N - 1037.5 N
= 1482.5 N
= 1.483 kN
Therefore, the forces on the right hand side is 1.483 kN
