Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J
Answer:
v₀ = 15 m/s
Explanation:
given,
initial velocity = v₀
down acceleration of rock = 10 m/s²
rock distance
S₄ = 7 x S₁
From kinematic equations
S = v₀ t+0.5 at²
at t = 1 s
S₁ = v₀ (1)+0.5 x 10 x 1²
S₁ = v₀+ 5......(1)
at t = 4 s
S₄ = v₀ (4)+0.5 x 10 x 4²
S₄ = 4 v₀+80.....(2)
from equation (1) and (2)
7( v₀ + 5 ) = 4 v₀ +80
3 v₀ = 80 - 35
3 v₀ = 45
v₀ = 15 m/s
Yes the velocity changes. Because velocity changes with direction. The object is moving around a gentle curve. The curve is not linear it is curve the direction changes a bit so obviously the velocity also changes but not much. Juts a minor change. Depends on how much curve the highway is.
<span>The lens system within a camera forms a real, inverted image
on the sensor.
Virtual images may be visible to us, but they don't register on
film or CCDs, because they're not real. And I think real images
are always inverted. (I'm not sure about that part.)
</span>
