The total quantity of electrons that have flowed through a circuit is a
quantity of charge, measured in Coulombs, or in Ampere-seconds.
The <em><u>rate</u></em> of flow of electrons, or more accurately the rate of flow of
the charge on them, is electrical current. Its unit is the Ampere.
1 Ampere is 1 Coulomb of charge per second.
CHECK COMPLETE QUESTION BELOW
you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth
Answer:
OPTION C is correct
The same as you do on Earth
Explanation :
According to law of gravitation :
F=GMm/R^2......(a)
F= mg.....(b)
M= mass of earth
m = mass of the person
R = radius of the earth
From law of motion
Put equation b into equation a
mg=GMm/R^2
g=GMm/R^2
g=GM/R^2
We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have
m= 4M
r=(2R)^2=4R^2
g= G4M/4R^2
Then, 4in the denominator will cancel out the numerator we have
g= GM/R^2
Therefore, g remain the same
Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.
Given Information:
Current = I = 20 A
Diameter = d = 0.205 cm = 0.00205 m
Length of wire = L = 1 m
Required Information:
Energy produced = P = ?
Answer:
P = 2.03 J/s
Explanation:
We know that power required in a wire is
P = I²R
and R = ρL/A
Where ρ is the resistivity of the copper wire 1.68x10⁻⁸ Ω.m
L is the length of the wire and A is the area of the cross-section and is given by
A = πr²
A = π(d/2)²
A = π(0.00205/2)²
A = 3.3x10⁻⁶ m²
R = ρL/A
R = 1.68x10⁻⁸*(1)/3.3x10⁻⁶
R = 5.09x10⁻³ Ω
P = I²R
P = (20)²*5.09x10⁻³
P = 2.03 Watts or P = 2.03 J/s
Therefore, 2.03 J/s of energy is produced in 1.00 m of 12-gauge copper wire carrying a current of 20 A
<span>the speed of a direction</span>
