Density = mass / volume
mass = 1.1 g
volume = length of side ^ 3 = [1.2 * 10^-5 km * 100000 cm/km]^3 = [1.2 cm]^3 = 1.728 cm^3
density = 1.1 g / 1.728 cm^3 = 0.64 g / cm^3
Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.
-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.
-- 'Watt' is a <u><em>rate</em></u> of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your <em>power</em> is 40.7 watts.
-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.
-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.
-- 'Kilowatt' is a bigger <em>rate</em> of using energy . . . 1,000 joules per second.
-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .
-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .
Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.
I attached a picture of the diagram associated with this question.
Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ
The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ
Answer:
0.799 m/s if air resistance is negligible.
Explanation:
For how long is the ball in the air?
Acceleration is constant. The change in the ball's height
depends on the square of the time:
,
where
is the change in the ball's height.
is the acceleration due to gravity.
is the time for which the ball is in the air.
is the initial vertical velocity of the ball.
- The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground.
. - Gravity pulls objects toward the earth, so
is also negative.
near the surface of the earth. - Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result,
.
Solve for
.
;
;
;
.
What's the initial horizontal velocity of the ball?
- Horizontal displacement of the ball:
; - Time taken:

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.
.
Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.