A ball thrown vertically upward reaches a maximum height of 30. meters above the surface of Earth. At its maximum height, the sp
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Answer:
13.78 m/s
Explanation:
Given that:
The mass of the girl & her bicycle = 42 kg
The speed at the top of the hilll= 4 m/s
The height of the hill = 14.3 m
The length of the hill (distance) = 112 m
The frictional force = 20 N
To find the speed at the bottom of the hill; we need to carry out the following processes.
The workdone by gravity = mass × acceleration due to gravity × Δh
= 42 × 9.8 × ( 14.3 - 0 )
= 5885.88 joules
The workdone by the friction = Force × distance = - 20 × 112 (since she is riding down the hill)
The workdone by the friction = -2240 Joules
The initial Kinetic friction = 1/2 mv²
= 1/2 × 42 × 4²
= 336 Joules
The final kinetic energy = Initial Kinetic energy + total work
The final kinetic energy = (336 + 5885.88 - 2240) Joules
The final kinetic energy = 3981.88 Joules
Using the final kinetic energy = 1/2 mv²
3981.88 = 1/2 × 42 × v²
3981.88 = 21 v²
v² = 3981.88/21
v² =189.61
v =
v = 13.78 m/s
Therefore, the speed at the bottom = 13.78 m/s
Answer:
<h2>40 kg</h2>Explanation:
Find the diagram relating to the question for proper explanation of the question below.
Using the principle of moment
Sum of clockwise moments = Sum of anticlockwise moments
Moment = Force * perpendicular distance
For anti-clockwise moment:
Since the 30 kg moves in the anticlockwise direction according to the diagram
ACW moment = 30 * 1 = 30 kgm
For clockwise moment
If another child sits 0.75 m away from the pivot point on the opposite side, moment of the child in clockwise direction = M * 0.75 = 0.75M (M is the mass of the unknown child).
Equating both moments we have;
0.75M = 30
M = 30/0.75
M = 40 kg
The second child's mass is 40 kg
