A driver in a 2144 kg car traveling at 15 m/s hits the brakes, coming to a stop in 67 meters. How far would it take the car to s
1 answer:
1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
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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
Answer:
200 C
Explanation:
Let C1 and C2 be their charges. According to Coulomb's law
where k = is the constant, R = 0.4m is the distance between them, F = 120 N is their resulting charge force
Since their total charge is 200C:
or
We can substitute the above equation
or
So the larger charge is C = 200 C