A 0.75 kg model car is moving west at a speed of 9.0 m/s when it collides head-on with a 2.00 kg model truck that is traveling e
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Answer:
The maximum speed of the car should be 13.7 m/s
Explanation:
For the car to travel at a maximum safe speed , the frictional force acting should be maximum and at the same time should provide the necessary centripetal force.
Let 'k' (=0.3502) be the coefficient of friction and 'N' be the normal force acting on the surface.
Then ,
N = mg , where 'm' is the mass of the body and 'g'(=9.8) is the acceleration due to gravity.
∴ Maximum frictional force , f = kN = kmg
Centripetal force that should act on the car to move with maximum possible speed is -
, where 'v' is the velocity of the car and 'r'(=55m) is the radius of circular path.
Equating the 2 forces , we get -
∴
Substituting all the values , we get -
v = 13.7 m/s.
Answer:
(a) Charge of 4.25 μF capacitor is 35.46 μC.
(b) Charge of 1.13 μF capacitor is 10.05 μC.
(c) Charge of 2.85 μF capacitor is 25.36 μC.
Explanation:
Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:
C₄ = C₁ + C₂
Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.
C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF
Since, C₄ and C₃ are connected in series, there equivalent capacitance is:
C₅ =
Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.
C₅ =
C₅ = 2.05 μF
The charge on the equivalent capacitance is determine by the relation :
Q = C₅ V
Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.
Q = 2.05 μF x 17.3 = 35.46 μC
Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.
Charge on the capacitor, C₃ = 35.46 μC
Charge on the capacitor, C₄ = 35.46 μC
Voltage on the capacitor C₄ = =
= 8.90 volts
Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.
Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC
Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC