Complete Question
A parallel plate capacitor creates a uniform electric field of 5 x 10^4 N/C and its plates are separated by 2 x 10^{-3}'m. A proton is placed at rest next to the positive plate and then released and moves toward the negative plate. When the proton arrives at the negative plate, what is its speed?
Answer:
Explanation:
From the question we are told that:
Electric field 
Distance 
At negative plate
Generally the equation for Velocity is mathematically given by
Therefore
Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
Answer:
Hey!
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Voltage (V) = 0.8V
Current (I) = 200 mA = 200/10^3 = 2/10
Resistance = ?
Resistance = Voltage / Current
Voltage = Current × Resistance
0.8 = 2/10 × Resistance
0.8×10/2 = Resistance
8/2 = Resistance
Resistance = 4 ohm
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Hope it helps...!!!
Explanation:
The atom is the most basic unit of matter
