Answer:
the thermal energy transferred is 183,226.68 J
Explanation:
Given;
mass of iron skillet, m = 5.1 kg
initial temperature, t₁ = 295 K
final temperature, t₂ = 373 K
The thermal energy transferred is calculated as;
Q = mcΔt
where;
Δt is change in temperature, = t₂ - t₁
c is specific heat capacity of iron = 460.6 J/kg.K
Q = 5.1 x 460.6 x (373 - 295)
Q = 183,226.68 J
Therefore, the thermal energy transferred is 183,226.68 J
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3
*cm^3 is a standard unit of volume*
<h3><u>Answer;</u></h3>
Doubles and Remains the same
<h3><u>Explanation;</u></h3>
- The effect of doubling the absolute temperature of a sample of a monoatomic ideal gas is that,the pressure doubles and density of the sample of gas remains the same.
- <em><u>According to ideal gas equation; PV = nRT; Where P is pressure and V is the Volume, n is the number of moles, R is the ideal gas constant and T is the absolute temperature.</u></em>
- <em><u>Therefore, when the temperature of the mono atomic ideal gas is doubled, the pressure of the gas will also doubles.</u></em>
- However, in a closed chamber mass of the ideal gas is invariant, since density depends only on the mass and volume therefore the density of the ideal is gas will remain the same.
Answer:
The smallest value is n= 2
Explanation:
The balmer equation is given below
1/λ = R(1/4 - 1/n₂²).
R= 1.0973731568508 × 10^7 m^-1
λ= 400*10^-9 m
(400*10^-9)= 1.0973731568508 × 10^7 (1/4-1/n²)
(400*10^-9)/1.0973731568508 × 10^7
= 1/4 - 1/n²
364.51 *10^-16= 1/4 - 1/n²
1/n²= 1/4 -364.51 *10^-16
1/n² = 0.25-3.6451*10^-14
1/0.25= n²
4= n²
√4= n
2= n
The smallest value is N= 2
To solve this problem we will apply the concept related to the electric field. The magnitude of each electric force with which a pair of determined charges at rest interacts has a relationship directly proportional to the product of the magnitude of both, but inversely proportional to the square of the segment that exists between them. Mathematically can be expressed as,
Here,
k = Coulomb's constant
V = Voltage
r = Distance
Replacing we have
Therefore the magnitude of the electric field is 
