Answer:
b)
Explanation:
If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.
For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)
For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)
Answer:
<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b)
The bullet travels horizontally 110.6 m</h2>
Explanation:
a) Consider the vertical motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Displacement, s = 1.5 m
Substituting
s = ut + 0.5 at²
1.5 = 0 x t + 0.5 x 9.81 xt²
t = 0.553 s
Time elapsed before the bullet hits the ground is 0.553 seconds.
b) Consider the horizontal motion of bullet
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 200 m/s
Acceleration, a = 0 m/s²
Time, t = 0.553 s
Substituting
s = ut + 0.5 at²
s = 200 x 0.553 + 0.5 x 0 x 0.553²
s = 110.6 m
The bullet travels horizontally 110.6 m
Answer:
Final momentum after a head on collision is -2kgm/s
Explanation:
One ball moves to the right and the other moves opposite and momentum is a vector quantity so that considering the direction
Initial momenta are P₁=2x3=6kgm/s P₂=4x(-2)=-8kgm/s
Final momentum is the vector sum of P(final)= 6-8= -2 kgm/s
