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Answer:
The second projectile was 1.41 times faster than the first.
Explanation:
In the ballistic pendulum experiment, the speed (v) of the projectile is given by:
<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum. </em>
To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:
(1)
<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively. </em>
Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):
Therefore, the second projectile was 1.41 times faster than the first.
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The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q =
Therefore, substitute the values into the above formula as follows.
Q =
=
=
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .