Answer:
Approximately
.
Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.
Explanation:
Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant
near the surface of the earth.
For an object that is accelerating constantly,
,
where
is the initial velocity of the object,
is the final velocity of the object.
is its acceleration, and
is its displacement.
In this case,
is the same as the change in the ball's height:
. By assumption, this ball was dropped with no initial velocity. As a result,
. Since the ball is accelerating due to gravity,
.
.
In this case,
would be the velocity of the ball just before it hits the ground. Solve for
.
.
Answer:
x(t) = -3sin2t
Explanation:
Given that
Spring force of, W = 720 N
Extension of the spring, s = 4 m
Attached mass to the spring, m = 45 kg
Velocity of, v = 6 m/s
The proper calculation is attached via the image below.
Final solution is x(t) = -3.sin2t
Answer:3,45 x 10^9 N
Explanation: We have considered the total charge for each coin , this is the total atoms x 29 electrons for cooper and multiplier by electron charge, the total charge for each coin is 0,464 C
Finally we use the Coulomb law,
F=k Q/ (r)^2
Answer:
because only two electrons can fit in the first orbit around the nucleus, and each period on the table is organized by number of orbits
Explanation:
Below is an attachment containing the solution.