Answer:
20.4e18 electrons/second ≈ 2e19 electrons/second
Explanation:
Hi!
To solve this problem we are going to use Omh's Law:
V = RI
And the relation ship between the resistance R and conductivity:
R = L/(σA)
*here we are considering a omhic material*
The conductance σ is related to electron mobility and electron density by:
σ = nμ
Replacing all these relations in the omhs law, we get:
V = (LI)/(σA)
We know that both wire are subject to the same electric field, therefore V is the same for both, moreover, since no additional info for the length of the wires is given we are going to consider that L is the same for both. Therefore
This means that:
From the relation of the conductance and electron mobility and density, and the data given to us, we know that:
Also
Therefore:
That is:
Since I_A = 3*10^18 e/s
I_B = 20.4*10^18 e/s
Answer:
0.0436 A
Explanation:
Current: This can be defined as the rate of flow of electric charge in a circuit. The S.I unit of current is Ampere (A).
The expression of current is given as
P = I²R
Where P = power, I = current, R = Resistance.
Make I the subject of the equation,
I = √(P/R)...................... Equation 1
Given: P = 0.25 W, R = 120 Ω
Substitute into equation 1
I = √(0.25/120)
I = √(0.00208)
I = 0.0456 A.
Hence the current that can safely pass through the resistor = 0.0436 A
Answer:
The density of the mercury is 13.2 g/cm³
Explanation:
Density is a measurement that compares the amount of matter an object
has to its volume
Density is equal to mass divided by volume
We need to find the density of mercury if 500 cm³ has a mass of
6.60 kg in g/cm
We must to change The kilogram to grams
The mass of mercury is 6.60 kilograms
1 kilogram = 1000 grams
6.60 kilograms = 6.60 × 1000 = 6600 grams
Density = mass ÷ volume
The volume of the mercury is 500 cm³
The density = 6600 ÷ 500
The density = 13.2 g/cm³
<em>The density of the mercury is 13.2 g/cm³ </em>
Answer:
0.784
Explanation:
Efficiency is the ratio of energy out over energy in.
We know the energy in was 1250 J.
The energy out is the change in potential energy of the box:
PE = mgh
PE = (25 kg) (9.8 m/s²) (4 m)
PE = 980 J
So the efficiency is:
e = 980 J / 1250 J
e = 0.784