Answer:
Part 1
Part 2
Part 3
Explanation:
Given
Number of protons
Radius of nucleus
Distance of the electrons
Part 1
Electric field produced by just outside its surface
Part 2
Electric field produced by just outside its surface
Part 3
The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other
hence, the solution is
Part 1
Part 2
Part 3
We know that According to Ohm's Law :
Current passing through a Conductor is directly proportional to the Voltage over a given Resistance.
⇒ V ∝ I
⇒ V = I × R
If Resistance is not changed and Voltage is increased, Based on Ohm's law we can conclude that Current flowing will also increase, because Voltage is directly proportional to Current.
Answer:
the final temperature of the tea is 7.39⁰C.
Explanation:
Given;
mass of the tea, m = 375 g
specific heat capacity of the tea, C = 4.184 JJ/g°C
initial temperature of the tea, t₁ = 95°C
the final temperature of the tea, t₂ = ?
Energy lost by the refrigerator, Q = 137,460 J
The energy lost by the refrigerator is given by the following formula;
-Q = mc(t₂ - t₁)
-137,460 =375 x 4.184(t₂ - 95°C)
-137,460 = 1569(t₂ - 95°C)
Therefore, the final temperature of the tea is 7.39⁰C.
Decrease the amount of work done.
Answer:
54%
Explanation:
So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.
Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.
Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.
The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:
2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.
Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.
Thus, the total percentage of the period does the mass lie in these regions = 54%.