Answer:a) 4+8+24=36
B) 1/4+1/8+1/24=10
C) yu will connect them in parallel connection.
D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.
Explanation:
Answer:
a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).
b) The centre of mass is still at the starting point for both bodies.
c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.
Explanation:
Linear momentum is conserved.
(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0
5×30 + 60 × v = 0
v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).
b) At time t = 10 s,
Sally has travelled 25 m and the boot has travelled 300 m.
Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.
Centre of mass = [(60)(25) + (5)(-300)]/(60+5)
= 0 m.
The centre of mass is still at the starting point for both bodies.
c) The shore is 30 m away.
Speed = (Distance)/(time)
Time = (Distance)/(speed) = (30/2.5)
Time = 12 s
Hope this Helps!!!
efficiency= [useful energy transferred ÷ total energy supply]×100%
So, [5500÷10000]×100%=0.55×100
=55%
The reading on the scale is the tension on the string that connects the two objecst. In order to support the blocks it must pull the weights by a force magnitude of W. So, the tension of the rope is W. Therefore, the reading on the scale is W, D.
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The correct answer is <span>3)

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In fact, the total energy of the rock when it <span>leaves the thrower's hand is the sum of the gravitational potential energy U and of the initial kinetic energy K:
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<span>As the rock falls down, its height h from the ground decreases, eventually reaching zero just before hitting the ground. This means that U, the potential energy just before hitting the ground, is zero, and the total final energy is just kinetic energy:
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But for the law of conservation of energy, the total final energy must be equal to the tinitial energy, so E is always the same. Therefore, the final kinetic energy must be
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