Answer:
281.25 J
Explanation:
We are told that the two objects with masses m and 3m.
Also that energy stored in the spring is 375 joules.
Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3
Thus, from principle of conservation of energy, we have;
½mv² + ½(3m)(v/3)² = 375J
(m + 3m/9)½v² = 375
(4/3)m × ½v² = 375
Multiply both sides by ¾ to get;
½mv² = 375 × ¾
½mv² = 281.25 J
Therefore, energy of lighter body is 281.25 J
Answer:
F = 50636.873 N
Explanation:
given,
bucket of water = 700-kg
length of cable = 20 m
Speed = 40 m/s
angle of the cable = 38.0°
let air resistance be = F
tension in rope be = T
T cos 38° = m×g..................(1)
..........(2)
equation (1)/(2)
F = 50636.873 N
Hence the force exerted on the bucket is equal to F = 50636.873 N
Answer:
Magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Given
Contact Time = t = 0.05 seconds
Mass (of ball) = 0.80kg
Initial Velocity = u = 25m/s
Final Velocity = 25m/s
Magnitude of the average force exerted on the wall by the ball is given by;
F = ma
Where m = 0.8kg
a = Average Acceleration
a = (u + v)/t
a = (25 + 25)/0.05
a = 50/0.05
a = 1000m/s²
Average Force = Mass * Average Acceleration
Average Force = 0.8kg * 1000m/s²
Average Force = 800kgm/s²
Average Force = 800N
Hence, the magnitude of the average force exerted on the wall by the ball is 800N
A) average acceleration = final velocity - initial velocity / time
= 7700 - 0 / 11
= 700ms^-2
B) force = mass x acceleration
= (3.05 x 105) x 700
= 320.25 x 700
= 224,175N
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