Answer:
The strength of the electric field is
.
Explanation:
Given that,
Speed 
Time 
Angle = 45°
We need to calculate the acceleration
Using equation of motion




We need to calculate the strength of the electric field
Using relation of newton's second law and electric force



Put the value into the formula


Hence, The strength of the electric field is
.
Answer:
The voltage is
Explanation:
From the question we are told that
The time that has passed is 
Here
is know as the time constant
The voltage of the power source is 
Generally the voltage equation for charging a capacitor is mathematically represented as
![V = V_b [1 - e^{- \frac{t}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7Bt%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\frac{\tau}{2}}{\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Cfrac%7B%5Ctau%7D%7B2%7D%7D%7B%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{\tau}{2\tau} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B%5Ctau%7D%7B2%5Ctau%7D%20%7D%5D)
=> ![V = V_b [1 - e^{- \frac{1}{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%20V_b%20%20%5B1%20-%20e%5E%7B-%20%5Cfrac%7B1%7D%7B2%7D%20%7D%5D)
=>
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval,

:

First we need to find the net charge flowing at a certain point of the wire in one second,

. Using I=0.92 A and re-arranging the previous equation, we find

Now we know that each electron carries a charge of

, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
EC_1 + EP_1 = EC2 + EP_2
EC_2 = 0
EC_2 = EP_1 - EP_2
EC_2 = mg(H_1 - H_2) = 0.20 kg * 9.8 m/s^2 * (3.25 m - 1.5m) = 3.43 J
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