Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
Here are the ones that I know about
and can think of just now:
-- wind
-- solar
-- nuclear
-- tidal
-- hydro
-- geothermal
-- biomass
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton,
= 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E

where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd




Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
I'm pretty sure it's D. negative acceleration