Answer:
3.25 × 10^7 m/s
Explanation:
Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)
Ek = 1/2 mv2 = qΔV .................. 1
Given that V is the electron speed in m/s
Charge of electron = 1.60217662 × 10-19 coulombs
Mass of electron = 9.109×10−31 kilograms
ΔV = 3.0kV = 3000V
Make V the subject of the formula in eqaution 1
V = sqr root 2qΔV/m
V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31
V = 3.25 × 10^7 m/s
The distance for any rectilinear motion at constant acceleration is:
d = v₀t + 0.5at²
where
v₀ is the initial velocity
So, if v₀ = 6v, and it stopped to 0 m/s, then the acceleration is equal to:
a = (0 - 6v)/t = -6v/t
Thus,
d = (6v)(t) + (0.5)(-6v/t)(t²)
d = 6vt - 3t
<span>d = 3t(2v - 1)</span>
Answer:
6.86 meters
Explanation:
Let the compression of the string be represented by x, and the height of projection of the toy rocket be represented by h.
So that;
x = 9 cm = 0.09 m
In its rest position (i.e before the launch), the spring has a stored potential energy which is given as;
PE =
K
=
x 830 x 
= 415 x 0.0081
= 3.3615
The potential energy in the string = 3.36 Joules
Also,
PE = mgh
where: m is the mass, g is the gravitational force and h the height.
m = 50 g = 0.05 kg, g = 9.8 m
Thus,
PE = 0.05 x 9.8 x h
3.3615 = 0.05 x 9.8 x h
3.3615 = 0.49h
⇒ h = 
= 6.8602
The height of the toy rocket would be 6.86 meters.
Answer:
i think option C is correct answer