Answer:
<em>There will be an increase in potential difference.</em>
Explanation:
As we know that the potential difference depends upon the capacitance.
ΔV = Q/C
When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.
In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.
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Answer:
The impulse exerted by one cart on the other has a magnitude of 4 N.s.
Explanation:
Given;
mass of the first cart, m₁ = 2 kg
initial speed of the first car, u₁ = 3 m/s
mass of the second cart, m₂ = 4 kg
initial speed of the second cart, u₂ = 0
Let the final speed of both carts = v, since they stick together after collision.
Apply the principle of conservation of momentum to determine v
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 3 + 0 = v(2 + 4)
6 = 6v
v = 1 m/s
Impulse is given by;
I = ft = mΔv = m(
The impulse exerted by the first cart on the second cart is given;
I = 2 (3 -1 )
I = 4 N.s
The impulse exerted by the second cart on the first cart is given;
I = 4(0-1)
I = - 4 N.s (equal in magnitude but opposite in direction to the impulse exerted by the first).
Therefore, the impulse exerted by one cart on the other has a magnitude of 4 N.s.
<span> One </span>volt<span> is </span>defined<span> as the difference in electric potential between two points of a conducting wire when an electric current of one ampere dissipates one watt of power between those points.</span>
