We use the following expression
T = 2*pi *sqrt(l/g)
Where T is the period of the pendulum
l is the length of the pendulum
and g the acceleration of gravity
We solve for l
l = [T/2*pi]² *g = [30s/2*pi]²* 9.8 [m/s²] = 223.413 m
The tower would need to be at least 223.413 m high
Answer:
Mass and volume.
Explanation:
The equation for density is always mass divided by volume. To determine the density of a fluid, you would need to find its volume and its mass.
Answer:
388.5J
Explanation:
Given parameters:
Weight = 70N
Height = 5.55m
Unknown:
Gravitational potential energy at the top of the ladder = ?
Solution:
The gravitational potential energy is the energy due to the position of the body.
Gravitational potential energy = Weight x height
So;
Gravitational potential energy = 70 x 5.55 = 388.5J
The difference in electric potential energy between the two points is

where q is the magnitude of the charge and

is the electric potential difference.
But for energy conservation, the difference in electric potential energy

between the two points is equal to the work done to move the charge between A and B:

so we have

and by substituting the numbers of the problem, we find the value of

:
Answer:

Explanation:
The acceleration of a circular motion is given by

where
is the angular velocity and
is the radius.
Angular velocity is related to the period, T, by

Substitute into the previous formula.


This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.