A force of 385 N is applied in pushing a stalled automobile at a constant speed for a distance of 150 m. How much work (in J) wa
1 answer:
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(a) 10241 W
In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.
The equation of the forces along the parallel direction is:
where
F is the force applied to pull the car
m = 950 kg is the mass of the car
is the acceleration of gravity
is the angle of the incline
Solving for F,
Now we know that the car is moving at constant velocity of
v = 2.20 m/s
So we can find the power done by the motor during the constant speed phase as
(b) 10624 W
The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation
where
v = 2.20 m/s is the final velocity
a is the acceleration
u = 0 is the initial velocity
t = 12.0 s is the time
Solving for a,
So now the equation of the forces along the direction parallel to the incline is
And solving for F, we find the maximum force applied by the motor:
The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:
(c)
Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.
The change in potential energy of the car is:
where
m = 950 kg is the mass
is the acceleration of gravity
is the change in height, which is
where L = 1250 m is the total distance covered.
Substituting, we find the energy transferred:
Answer:
a) the capacitance is the same for both capacitors.
b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q.
c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q
d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q
Explanation:
a) The capacitance of a capacitor, by definition, is as follows:
Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be expressed as follows:
C = ε*A / d
As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.
b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.
If this proportion must remain the same, and one of the capacitors has the double of the charge than the other, the potential difference must be the double also.
c) Applying Gauss' law, to the surface of one of the plates, and assuming a constant surface charge density σ, it can be showed that the electric field can be calculated as follows:
E*A = Q/ε₀ as σ=Q/A
⇒ E = σ/ε₀
As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.
c) The electric potential energy, stored between plates of a capacitor, can be written as follows:
If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.