Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the relationship between the tension in the string and the gravitational force acting on the pendulum bob?
A.T = mg
B.T = mg cos(θ)
C.T = mg sin(θ)
D.T = mg tan(θ)
Solution
Tension provides the vertical component to balance the weight: T cos(θ) = mg.
Q. A conical pendulum swings in a horizontal circle. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?
A.T = mg
B.T = mg/cos(θ)
C.T = mg/sin(θ)
D.T = mg tan(θ)
Solution
The vertical component of tension balances the weight: T cos(θ) = mg, thus T = mg/cos(θ).
Q. A conical pendulum swings with a constant speed. If the angle of the string with the vertical is θ, what is the expression for the tension in the string?
A.mg/cos(θ)
B.mg/sin(θ)
C.mg/tan(θ)
D.mg
Solution
Tension T = mg/cos(θ) to balance the vertical component of weight.
Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 60 cm from the lens, what is the distance of the image from the lens?
A.15 cm
B.30 cm
C.45 cm
D.60 cm
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 30 cm.
Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 40 cm from the lens, what is the distance of the image from the lens?
A.20 cm
B.40 cm
C.60 cm
D.80 cm
Solution
Using the lens formula 1/f = 1/v - 1/u, where f = 20 cm and u = -40 cm, we find v = 20 cm. The image is formed at 20 cm on the opposite side.
Q. A convex lens has a focal length of 20 cm. If an object is placed at a distance of 30 cm from the lens, what is the distance of the image from the lens?
A.60 cm
B.15 cm
C.30 cm
D.10 cm
Solution
Using the lens formula 1/f = 1/v - 1/u, we find v = 60 cm.
Q. A cyclist accelerates from rest to a speed of 15 m/s. If the mass of the cyclist and the bicycle is 75 kg, what is the kinetic energy at that speed?
A.500 J
B.750 J
C.1000 J
D.1250 J
Solution
Kinetic Energy = 0.5 × mass × velocity² = 0.5 × 75 kg × (15 m/s)² = 8437.5 J.
Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the speed of the cyclist relative to the pedestrian?
A.10 m/s
B.15 m/s
C.5 m/s
D.20 m/s
Solution
Relative speed = Speed of cyclist - Speed of pedestrian = 15 m/s - 5 m/s = 10 m/s.
Q. A cyclist is moving at 15 m/s and a pedestrian is walking at 5 m/s in the same direction. What is the relative speed of the pedestrian with respect to the cyclist?
A.10 m/s
B.5 m/s
C.20 m/s
D.15 m/s
Solution
Relative speed = Speed of pedestrian - Speed of cyclist = 5 m/s - 15 m/s = -10 m/s (10 m/s behind).
Q. A cyclist is moving at 15 m/s and passes a stationary observer. If the observer starts moving at 5 m/s in the same direction, what is the speed of the cyclist relative to the observer?
A.10 m/s
B.15 m/s
C.20 m/s
D.5 m/s
Solution
Relative speed = Speed of cyclist - Speed of observer = 15 m/s - 5 m/s = 10 m/s.
Q. A cyclist is moving at 15 m/s towards the east while a car is moving at 25 m/s towards the west. What is the relative speed of the cyclist with respect to the car?
A.10 m/s
B.15 m/s
C.40 m/s
D.25 m/s
Solution
Relative speed = Speed of cyclist + Speed of car = 15 m/s + 25 m/s = 40 m/s.