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Ede4ka [16]
3 years ago
14

While pushing a refrigerator up a ramp, there would be less friction to overcome if you used a dolly because of the wheels. So t

he efficiency of the ramp would be ________________________ with the use of the dolly.
Physics
1 answer:
FromTheMoon [43]3 years ago
4 0

While pushing a refrigerator up a ramp, there would be less friction to overcome if you used a dolly because of the wheels. So the efficiency of the ramp would be greater with the use of the dolly.

<u>Explanation:</u>

The ramp provides for a wedge setup, so when the refrigerator is being pushed up the ramp, the friction on the ramp would equally distributed to the refrigerator when pushed wholly as the friction applied is a product of the coefficient of friction with the Normal reaction of the weight in contact with the surface.

Therefore, when the dolly is used, the surface in contact becomes divided and thus, the friction applied would become less due to though coefficient of friction remains the same, the normal reaction of the weight becomes small and so does the applied friction. Therefore, efficiency of the ramp increases as the friction is less and work done is more.

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Answer:

(a) Magnitude: 14.4 N

(b) Away from the +6 µC charge

Explanation:

As the test charge has the same sign, the force that the other charges exert on it will be a repulsive force. The magnitude of each of the forces will be:

F_e = K\frac{qq_{test}}{r^2}

K is the Coulomb constant equal to 9*10^9 N*m^2/C^2, q and qtest is the charge of the particles, and r is the distance between the particles.

Let's say that a force that goes toward the +6 µC charge is positive, then:

F_e_1 = K\frac{q_1q_{test}}{r^2}=-9*10^9 \frac{Nm^2}{C^2} \frac{6*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =-21.6 N

F_e_2 = K\frac{q_2q_{test}}{r^2}=9*10^9 \frac{Nm^2}{C^2} \frac{2*10^{-6}C*4*10^{-6}C}{(0.1m)^2} =7.2 N

The magnitude will be:

F_e = -21.6 + 7.2 = -14.4 N, away from the +6 µC charge

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