Which of the following forces exists between objects even in the absence of direct physical contact
2 answers:
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Complete Question
For each of the following scenarios, describe the force providing the centripetal force for the motion:
a. a car making a turn
b. a child swinging around a pole
c. a person sitting on a bench facing the center of a carousel
d. a rock swinging on a string
e. the Earth orbiting the Sun.
Answer:
Considering a
The force providing the centripetal force is the frictional force on the tires \
i.e
where is the coefficient of static friction
Considering b
The force providing the centripetal force is the force experienced by the boys hand on the pole
Considering c
The force providing the centripetal force is the normal from the bench due to the boys weight
Considering d
The force providing the centripetal force is the tension on the string
Considering e
The force providing the centripetal force is the force of gravity between the earth and the sun
Explanation:
Answer:
The elastic potential energy is zero.
The net force acting on the spring is zero.
Explanation:
The equilibrium position of a spring is the position that the spring has when its neither compressed nor stretched - it is also called natural length of the spring.
Let's now analyze the different statements:
The spring constant is zero. --> false. The spring constant is never zero.
The elastic potential energy is at a maximum --> false. The elastic potential energy of a spring is given by
where k is the spring constant and x the displacement. Therefore, the elastic potential energy is maximum when x, the displacement, is maximum.
The elastic potential energy is zero. --> true. As we saw from the equation above, the elastic potential energy is zero when the displacement is zero (at the equilibrium position).
The displacement of the spring is at a maxi num --> false, for what we said above
The net force acting on the spring is zero. --> true, as the spring is neither compressed nor stretched
Answer:
160 Hz , 240 Hz , 400 Hz
Explanation:
Given that
Frequency of forth harmonic is 320 Hz.
Lets take fundamental frequency = f₁
f₁=80 Hz
Frequency of first harmonic = f₂
f₂=2 f₁
f₂ =2 x 80 = 160 Hz
Frequency of second harmonic = f₃
f₃= 3 f₁=3 x 80 = 240 Hz
Frequency of fifth harmonic = f₅
f₅= 5 f₁= 5 x 80 = 400 Hz
Three frequencies are as follows
160 Hz , 240 Hz , 400 Hz