A simple pendulum consists of a brass sphere of mass m = 1 kg suspended on a string of length L ≈ 1 meter. The pendulum oscillat
1 answer:
Answer:
2. Using a shorter string of length L ′ ≈ 0.25 meters
5. Using a shorter string of length L ′ ≈ 0.5 meters
Explanation:
The period of a pendulum is given by
where
L is the length of the pendulum
g is the acceleration due to gravity
We see from the formula that the period of the pendulum depends only on its length, not on its mass or its amplitude of ocillation. Therefore, the only alterations that can change the period of the pendulum are the ones where its length is changed.
Moreover, we notice that the period is proportional to the square of the length: this means that in order to decrease the period of the pendulum (the problem asks us which alterations will reduce the period of the pendulum from 2 s to 1 s), the length of the pendulum should also be reduced.
Therefore, the only alterations that will reduce the period of the pendulum are:
2. Using a shorter string of length L ′ ≈ 0.25 meters
5. Using a shorter string of length L ′ ≈ 0.5 meters
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Answer:
15.1°
Explanation:
The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:
Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:
(1)
where
is the initial vertical velocity
g = 9.8 m/s^2 is the gravitational acceleration
t is the time
Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by
Substituting t into (1) we find the final vertical velocity
where the negative sign means that the velocity is downward.
Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know from the Coulomb's Law that, Coulomb's force is directly proportional to the product of two charges q1 and q2 and inversely proportional to the square of the radius between them.
So,
F =
Now, we are asked to get the greatest force. So, in order to do that, product of the charges must be greatest because the force and product of charges are directly proportional.
Let's suppose, q1 = q
So,
if q1 = q
then
q2 = Q-q
Product of Charges = q1 x q2
Now, it is:
Product of Charges = q x (Q-q)
So,
Product of Charges = qQ -
And the expression qQ - is clearly a quadratic expression. And clearly its roots are 0 and Q.
So, the highest value of the quadratic equation will be surely at mid-point between the two roots 0 and Q.
So, the midpoint is:
q =
q = Q/2 and it is the highest value of each charge in order to get the greatest force.