Answer:
8.75
Explanation:
First, find the force of friction.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.9 m/s)² = F (1.4 m)
F = 11.7 N
Next, find the distance at the new velocity.
Kinetic energy = work done by friction
½ mv² = Fd
½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d
d = 8.75 m
Answer:
The energy in its ground state is 10 meV.
Explanation:
It is given that,
The energy of the electron in its first excited state is 40 meV.
Energy of the electron in any state is given by :
For ground state, n = 1
.............(1)
For first excited state, n = 2
.............(2)
Dividing equation (1) and (2), we get :
So, the energy in its ground state is 10 meV. Hence, this is the required solution.
K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2
So, option B is your answer!
Hope this helps!
Answer:The anwser is B i Promise ok
Explanation:
