Answer:
v = 4.8 10⁻⁴ m / s
Explanation:
To solve this exercise we can use the concepts of energy. In this case the potential energy is transformed into kinetic energy
U = K
q V = ½ m v²
v = [tex]\sqrt { \frac{2qV}{m} }[/tex]
in the exercise they indicate the value of the charge q₁ = 1 pC = 1 10⁻¹² C
let's calculate
v = [tex]\sqrt{ \frac{2 \ 1 \ 10^{-12} 120 \ 10^3}{500 ^{2} }[/tex]
v = 4.8 10⁻⁴ m / s
1 ) when a ball is projected upwords its time of rising is ...............the time of falling .
a) greater than b) smaller than c) equal to d ) double
2 ) when an object falls freely under the effect of gravity , the distance moved is
a ) directly proportional to time
b ) inversely proportional to time
c ) directly proportional to square of time
d ) inversely proportional to square of time.
Answer:
correct answer is C
Explanation:
In this exercise, you are asked to complete the sentences so that the sentence makes sense.
1) in projectile launching, the only force that acts is gravity in the vertical direction, so the time of going up is EQUAL to the time of going down
correct answer C
2) when a body falls freely, the acceleration is the ratio of gravity, therefore if it starts from rest, its height is
y = v₀ t - ½ gt²
v₀ = 0
y = -1/2 g t²
so the position is not proportional to the square of the time
correct answer is C