Answer:
Explanation:
Here image distance is fixed .
In the first case if v be image distance
1 / v - 1 / -25 = 1 / .05
1 / v = 1 / .05 - 1 / 25
= 20 - .04 = 19.96
v = .0501 m = 5.01 cm
In the second case
u = 4 ,
1 / v - 1 / - 4 = 1 / .05
1 / v = 20 - 1 / 4 = 19.75
v = .0506 = 5.06 cm
So lens must be moved forward by 5.06 - 5.01 = .05 cm ( away from film )
That statement is physically and grammatically true but legally false.
The impulse exerted on the ball by the wall is 24i kgm/s.
<h3>
Impulse exerted by the ball on the wall</h3>
The impulse exerted by the ball on the wall is the change in the linear momentum of the ball.
J = ΔP
ΔP = Pf - Pi
P = mv
where;
- m is mass of the ball
- v is the velocity
ΔP = 3(4.0i + 3.0j) - 3(-4.0i + 3.0j)
ΔP = (12i + 9j) - (-12i - 9j)
ΔP = 24i kgm/s
Thus, the impulse exerted on the ball by the wall is 24i kgm/s.
Learn more about impulse here: brainly.com/question/904448
#SPJ1
1.) It will accelerate "Leftwards"
2.) Momentum (p) = mv = 10*10 = 100m/s
So, your final answer is 100 m/s
Hope this helps!
<span>my first part it like that as we all know that flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2
</span><span>NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2
calculate required region area with formula =double integral(r^2sintheta dtheta dphi)
we get =.106m^2 now multiply D*Required region we get 7.5uc</span>