Q. A box is pushed across a floor with a force of 50 N. If the coefficient of kinetic friction is 0.4, what is the net force acting on the box if the normal force is 100 N?
A.10 N
B.20 N
C.30 N
D.40 N
Solution
Frictional force = μk * N = 0.4 * 100 N = 40 N. Net force = applied force - frictional force = 50 N - 40 N = 10 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.3. If the normal force is 100 N, what is the net force acting on the box?
A.20 N
B.30 N
C.50 N
D.70 N
Solution
Frictional force = μk * N = 0.3 * 100 N = 30 N. Net force = applied force - frictional force = 50 N - 30 N = 20 N.
Q. A box is pushed with a force of 50 N on a surface with a coefficient of kinetic friction of 0.4. What is the acceleration of the box if its mass is 10 kg?
A.1 m/s²
B.2 m/s²
C.3 m/s²
D.4 m/s²
Solution
Net force = applied force - frictional force. Frictional force = μ_k * N = 0.4 * 10 kg * 9.8 m/s² = 39.2 N. Net force = 50 N - 39.2 N = 10.8 N. Acceleration = F/m = 10.8 N / 10 kg = 1.08 m/s², approximately 1 m/s².
Q. A building is 40 m high. From a point on the ground, the angle of elevation to the top of the building is 60 degrees. What is the distance from the point to the base of the building?
A.20√3 m
B.40 m
C.30 m
D.10√3 m
Solution
Using tan(60°) = height/distance, we have distance = height/tan(60°) = 40/√3 = 20√3 m.
Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. If the frequency of the AC source is increased, what happens to the capacitive reactance?
A.Increases
B.Decreases
C.Remains the same
D.Becomes infinite
Solution
Capacitive reactance (X_C) is given by X_C = 1/(2πfC). If the frequency (f) increases, X_C decreases.
Q. A capacitor in an AC circuit has a capacitive reactance of 50 ohms. What is the frequency if the capacitance is 10 microfarads?
A.1 kHz
B.10 kHz
C.100 Hz
D.1000 Hz
Solution
Capacitive reactance (X_C) is given by X_C = 1 / (2πfC). Rearranging gives f = 1 / (2πX_CC). Substituting X_C = 50 ohms and C = 10 x 10^-6 F gives f = 318.31 Hz, approximately 1 kHz.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is doubled, what happens to the voltage across the capacitor?
A.It doubles
B.It halves
C.It remains the same
D.It quadruples
Solution
When the distance is doubled, the capacitance decreases, leading to an increase in voltage since Q = CV is constant.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. If the distance between the plates is increased, what happens to the charge?
A.Increases
B.Decreases
C.Remains the same
D.Becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not affect the charge.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the distance between the plates is increased?
A.Charge increases
B.Charge decreases
C.Charge remains the same
D.Charge becomes zero
Solution
When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not change the charge.
Q. A capacitor is charged to a voltage V and then disconnected from the battery. What happens to the charge on the capacitor if the voltage is doubled?
A.Charge doubles
B.Charge halves
C.Charge remains the same
D.Charge quadruples
Solution
The charge on a capacitor is given by Q = C * V. If the voltage is doubled, the charge also doubles, assuming capacitance remains constant.
