Answer and Explanation:
This can be explained as in Rutherford's model of atom the electrons orbits the nucleus which means that they will travel around the nucleus with some velocity and hence radiate electromagnetic waves which results in the loss of energy due to which the electron keeps coming closer and eventually falls into the nucleus.
But Bohr came up with a better explanation as according to the Bohr's atomic model, electrons stay fixed in orbit with certain energy in different shells around the nucleus and can only jump from an energy level to another if that specific amount of energy is supplied to it.
This model is based on the quantization of energy thus giving an explanation why electrons do not fall into the nucleus of an atom.
Answer:
300 K
Explanation:
First, we have find the specific heat capacity of the unknown substance.
The heat gained by the substance is given by the formula:
H = m*c*(T2 - T1)
Where m = mass of the substance
c = specific heat capacity
T2 = final temperature
T1 = initial temperature
From the question:
H = 200J
m = 4 kg
T1 = 200K
T2 = 240 K
Therefore:
200 = 4 * c * (240 - 200)
200 = 4 * c * 40
200 = 160 * c
c = 200/160
c = 1.25 J/kgK
The heat capacity of the substance is 1.25 J/kgK.
If 300 J of heat is added, the new heat becomes 500 J.
Hence, we need to find the final temperature, T2, when heat is 500 J.
Using the same formula:
500 = 4 * 1.25 * (T2 - 200)
500 = 5 * (T2 - 200)
100 = T2 - 200
=> T2 = 100 + 200 = 300 K
The new final temperature of the unknown substance is 300K.
Heat lost or gained, H = mc(θ₂ - θ₁)
Where m = mass, c = Specific heat capacity, θ₂= final temperature, θ₁ = initial temperature
m = 200g, c = 0.444 J/g°C, θ₁ = 22 °C (Since it was cooled).
H = 6.9 kj = 6.9 *1000J = 6900 J
6900 = 200*0.444* (θ₂ - 22)
6900/(200*0.444) = θ₂ - 22
77.70 = θ₂ - 22
θ₂ - 22 = 77.7
θ₂ = 77.7 + 22 = 99.7
So initial temperature before cooling ≈ 100°C . Option C.
Answer:
the average force the blade exerts on the log is 1791.05 N.
Explanation:
Given;
mass of the ax head, m = 4 kg
speed of the ax, v = 3 m/s
depth traveled into the log, d = 0.01 m
The time to traveled through the depth;
The average force the blade exerts on the log;
Therefore, the average force the blade exerts on the log is 1791.05 N.
