Answer:
2.5 m/s east
Explanation:
Let east be the positive direction for velocity.
The change in momentum of the 0.75 kg model car is ...
m1·v2 -m1·v1 = (0.75 kg)(11 m/s) -(0.75 kg)(-9 m/s)
= (0.75 kg)(20 m/s) = 15 kg·m/s
The change in momentum of the 2.0 kg model car is the opposite of this, so the total change in momentum is zero.
m2·v2 -m2·v1 = (2 kg)(v2 m/s) -(2 kg)(10 m/s) = 2(v2 -10) kg·m/s
The required relation is ...
15 kg·m/s = -2(v2 -10) kg·m/s
-7.5 = v2 -10 . . . . divide by -2
2.5 = v2 . . . . . . . add 10
The velocity of the model truck after the collision is 2.5 m/s east.
Answer:
(a). Energy is 64,680 J
(b) velocity is 51.43m/s
(c) velocity in mph is 115.0mph
Explanation:
(a).
The potential energy 
of the payload of mass 
is at a vertical distance 
is
.
Therefore, for the payload of mass 
at a vertical distance of 
, the potential energy is
(b).
When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,
(c).
The velocity in mph is
Runner 2 sees Runner 1 passing him with a velocity of 17 m/s west.
Given Information:
Voltage of circuit A = Va = 208 Volts
Current of circuit A = Ia = 40 Amps
Voltage of circuit B = Vb = 120 Volts
Current of circuit B = Ib = 20 Amps
Required Information:
Ratio of power = Pa/Pb = ?
Answer:
Ratio of power = Pa/Pb = 52/15
Explanation:
Power can be calculated using Ohm's law
P = VI
Where V is the voltage and I is the current flowing in the circuit.
The power delivered by circuit A is
Pa = Va*Ia
Pa = 208*40
Pa = 8320 Watts
The power delivered by circuit B is
Pb = Vb*Ib
Pb = 120*20
Pb = 2400 Watts
Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is
Pa/Pb = 8320/2400
Pa/Pb = 52/15
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