Answer:
I = Δq / t
Explanation:
The quantity of electricity i.e charge is related to current and time according to the equation equation:
Q = It
Δq = It
Where:
Q => is the quantity of electricity i.e charge
I => is the current.
t => is the time.
Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:
Δq = It
Divide both side by t
I = Δq / t
The energy of a photon is given by

where

is the Planck constant
f is the frequency of the photon
In our problem, the frequency of the light is

therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
~Formula: Voltage= current• resistance
(V= Ir)
~Using this formula, plug in the numbers from the equation into the formula
~5=25i
~Now you have a one-step equation
~Divide by 25 on both sides and you should get your answer:
~I= 0.2 (which means current is 0.2)
The second object, the one that had twice the force applied to it, would move twice as far, I believe.
Answer:

Explanation:
When an amount of energy Q is supplied to a substance of mass m, the temperature of the substance increases by
, according to the equation

where
is the specific heat capacity of the substance.
In this problem, we have:
is the amount of heat supplied to the sample of gold
m = 0.1 kg = 100 g is the mass of the sample
is the specific heat capacity of gold
Solving for
, we find the change in temperature

And since the final temperature was

The initial temperature was
