A vehicle moving at 5m/s, i . what should be the constant declaration in order to stop it within 15m? ii. How long it takes to s
1 answer:
Answer:
Refer to the attachment for solution (1).
<h3><u>Calculating time taken by it to stop (t) :</u></h3>By using the second equation of motion,
→ v = u + at
- v denotes final velocity
- u denotes initial velocity
- t denotes time
- a denotes acceleration
→ 0 = 5 + (-5/6)t
→ 0 = 5 - (5/6)t
→ 0 + (5/6)t = 5
→ (5/6)t = 5
→ t = 5 ÷ (5/6)
→ t = 5 × (6/5)
→ t = 6 seconds
→ Time taken to stop = 6 seconds
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Answer:
The potential difference between the plates increases
Explanation:
As we know that the capacitance of the capacitor is given by:
(1)
where
q = charge
C = capacitance
V = Voltage or Potential Difference
Also, the capacitance of a parallel plate capacitor is given as:
(2)
where
A = Area of the plates
D = Separation distance between the plates
Now, from eqn (1) and (2):
Now, from the above eqn we can say that:
Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.
Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.