Answer:
The answer is B: to Amplify the sound
Protons do not move out of the nucleus of atoms although they repel each other.
Remember that protons are particles with positive charge and they held together in the nucleus of the atom which is a tiny tiny region. As you know, like charges repel each other, which means that the protons exert a repulsion force.
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¢.
We use v=IR and assuming the resistance doesn’t change we can also say that the voltage and current (I) are directly proportional which means the voltage also decreases by 1/2
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
