A diver is on a board 1.80 m above
2 answers:
Answer:
v = 6.95 m/s
Explanation:
Given that,
A diver is on a board 1.80 m above the water, s = 1.8 m
The initial speed of the diver, u = 3.62 m/s
Let v is the speed with which she hit the water. It will move under the action of gravity. Using the equation of motion as follows :
So, she will hit the water with a speed of 6.95 m/s.
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Answer:
Number of electrons are flowing per second is 2.42 x 10¹⁹
Explanation:
The electric current flows through a wire is given by the relation :
....(1)
Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.
But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,
....(2)
Here N is the rate of electrons passing through junction.
From equation (1) and (2).
But area of wire,
Here d is diameter of wire.
So,
Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.
N = 2.42 x 10¹⁹ s⁻¹
Answer:
Stay the same
Explanation:
First of all, let's find how the capacitance of the capacitor changes.
Initially, it is given by
where
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:
The potential difference across the capacitor is given by
where
Q is the charge on the plates
C is the capacitance
We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:
Now we can analyze the electric field between the plates of the capacitor, which is given by
we said that:
- The voltage has doubled: V' = 2 V
- The distance between the plates has doubled: d' = 2 d
therefore, the new electric field will be
So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:
F = qE
Then the force on the particle will stay the same.