Kinetic Energy. Energy is transferred from one object to another when a reaction takes place. Energy comes in many forms and can be transferred from one object to another as heat, light, or motion, to name a few. For the blue ball to move to the position of the green ball, energy must be given to the blue ball.
Answer:
d) 2Fr
Explanation:
We know that the work done in moving the charge from the right side to the left side in the k shell is W = ∫Fdr from r = +r to -r. F = force of attraction between nucleus and electron on k shell. F = qq'/4πε₀r² where q =charge on electron in k shell -e and q' = charge on nucleus = +e. So, F = -e × +e/4πε₀r² = -e²/4πε₀r².
We now evaluate the integral from r = +r to -r
W = ∫Fdr
= ∫(-e²/4πε₀r²)dr
= -∫e²dr/4πε₀r²
= -e²/4πε₀∫dr/r²
= -e²/4πε₀ × -[1/r] from r = +r to -r
W = e²/4πε₀[1/-r - 1/+r] = e²/4πε₀[-2/r} = -2e²/4πε₀r.
Since F = -e²/4πε₀r², Fr = = -e²/4πε₀r² × r = = -e²/4πε₀r and 2Fr = -2e²/4πε₀r.
So W = -2e²/4πε₀r = 2Fr.
So, the amount of work done to bring an electron (q = −e) from right side of hydrogen nucleus to left side in the k shell is W = 2Fr
Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:
Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ = 
A₂ = Area at the throat
A₂ = 
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or
or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Answer:
–77867 m/s/s.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:
Acceleration = (final velocity – Initial velocity) /time
a = (v – u) / t
With the above formula, we can obtain acceleration of the ball as follow:
Initial velocity (u) = 34.5 m/s
Final velocity (v) = –23.9 m/s
Time (t) = 0.00075 s
Acceleration (a) =?
a = (v – u) / t
a = (–23.9 – 34.5) / 0.00075
a = –58.4 / 0.00075
a = –77867 m/s/s
Thus, the acceleration of the ball is –77867 m/s/s.
