Answer:
it have Potential energy
Explanation:
given data
Drag the pendulum to an angle 30∘
to find out
what form of energy does it have
solution
we know that pendulum start no kinetic energy when it release from any rest position then in starting it have potential energy only so that when pendulum is angle 30∘ at some height from ground so when it start it have potential energy same as in starting.
we know that the total energy is always conserve
so it have potential energy
Answer:
0.23 s
Explanation:
First of all, let's find the time constant of the circuit:

where
is the resistance
is the capacitance
Substituting,

The charge on a charging capacitor is given by
(1)
where
is the full charge
we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

Substituting this into eq.(1) we find

Answer:
A. False
B True
C. False
D.False
E. True
F. False
G. False
H. False
I. True
Explanation:
A. False: The system being analyzed consists of the bug and the car. These are the two bodies involved in the collision.
B. True: The system being analyzed consists of the bug and the car
C. False: The magnitudes of the change in velocity are different from the car and the bug. The velocity of the bug changes from 0 to the velocity of the car, while there is no noticeable change in the velocity of the car
D.False: There is barely any change in the momentum of the car since the mass of the bug is very small.
E. True: Since the mass of the bug is small, and was initially at rest, the magnitude of the change in monentum will be large because the new velocity will be that of the car.
F. False: The system being analyzed consists of the bug and the car. Those are the two bodies involved in the collision
G. False: The car barely changes in velocity since the mass of the bug is small.
H. False: The car barely changes in momentum because the collision does not affect its speed so much. on the other hand the momentum change of the bug is large since its mass is small.
I. True: The bug which was initially at rest will begin moving with the velovity of the speeding car, while the car barely changes in its velocity
Answer:
To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement
When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses
We can also use the equation and multiply the constant K by the displacement and this is the applied weight.
Explanation:
For this problem we will use the translational equilibrium relation
F –W = 0
F = W
W = mg
The spring elastic force is
F = - k x
We substitute
k x = m g
Where we see that the force of the spring is equal to the weight of the body.
To find the value of the unknown weight, we previously placed the 3 known weights and made a graph of the force against displacement
When hanging the weight is known, we measure the displacement and from the graph we can find the value of the hanging masses
We can also use the equation and multiply the constant K by the displacement and this is the applied weight.