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
Answer:
3 years
Explanation:
Find the circumference of each orbit in AU.
2xπx1=6.283185307
2xπx3=18.84955592
Divide them.
18.84955592/6.283185307=3
3 years
Answer:
Time period, 
Explanation:
Given that,
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz
We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :
T is the time period of the crystal's motion.
Time period is given by :
So, the time period of the crystal's motion is 
. Hence, this is the required solution.
Solution:
54 / 9 = 6 boxes.
Answer:
v = 2.928 10³ m / s
Explanation:
For this exercise we use Newton's second law where the force is the gravitational pull force
F = ma
a = F / m
Acceleration is
a = dv / dt
a = dv / dr dr / dt
a = dv / dr v
v dv = a dr
We substitute
v dv = a dr
∫ v dv = 1 / m G m M ∫ 1 / r² dr
We integrate
½ v² = G M (-1 / r)
We evaluate from the lower limit v = 0 for r = R m to the upper limit v = v for r = R + 2.73 10³, where R is the radius of Saturn's moon
v² = 2G M (- 1 / R +2.73 10³+ 1 / R)
We calculate
v² = 2 6,674 10⁻¹¹ 1.10 10²¹ (10⁻³ / 5.61 - 10⁻³ /(5.61 + 2.73))
v² = 14.6828 10⁷ (0.1783 -0.1199)
v = √8.5748 10⁶
v = 2.928 10³ m / s
