Answer:
The speed of proton when it emerges through the hole in the positive plate is
.
Explanation:
Given that,
A parallel-plate capacitor is held at a potential difference of 250 V.
A A proton is fired toward a small hole in the negative plate with a speed of, 
We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :

So, the speed of proton when it emerges through the hole in the positive plate is
.
Answer:
initial velocity is v = 4.95 m / s
Explanation:
To solve this exercise we use the projectile launch ratios, when the block leaves the hill its speed is horizontal, let's find the time it takes to fall to the other point.
Initial vertical velocity is zero
y = y₀ + v_{oy} t - ½ g t²
y-y₀ = 0 -1/2 g t²
t = 
calculate
t =
t = 2.02 s
with this time we can substitute in the horizontal displacement equation
x = v₀ₓ t
v₀ₓ = x / t
suppose that the distance between the two points is x = 10 m
v₀ₓ = 10 / 2.02
v₀ₓ = 4.95 m / s
initial velocity is v = 4.95 m / s
Answer:
The take-off speed is 41.48 
Explanation:
Given :
Range
m
Projectile angle
13°
From the formula of range,

Find the velocity from above equation,

( ∵
)

Therefore, the take-off speed is 41.48
First we have to establish that the number of protons is equivalent to the atomic number of element. Here I am assuming that you are referring to Potassium (K) - 40. Potassium, stable or unstable has 19 protons.