We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
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Explanation:
I want to say option B - Both forces can act without objects touching.
Answer:
x_total = (A + B) cos (wt + Ф)
we have the sum of the two waves in a phase movement
Explanation:
In this case we can see that the first boy Max when he enters the trampoline and jumps creates a harmonic movement, with a given frequency. When the second boy Jimmy enters the trampoline and begins to jump he also creates a harmonic movement. If the frequency of the two movements is the same and they are in phase we have a resonant process, where the amplitude of the movement increases significantly.
Max
x₁ = A cos (wt + Ф)
Jimmy
x₂ = B cos (wt + Ф)
total movement
x_total = (A + B) cos (wt + Ф)
Therefore we have the sum of the two waves in a phase movement
Answer:
<h2>The pin's final velocity is 5m/s</h2>
Explanation:
Step one:
given data
mass of ball m1=5kg
initial velocity of ball u1=10m/s
mass of pin m2=2kg
initial velocity of pin u2= 0m/s
final velocity of ball v2=8m/s
final velocity of pin v2=?
Step two:
The expression for elastic collision is given as
m1u1+m2u2=m1v1+m2v2
substituting we have
5*10+2*0=5*8+2*v2
50+0=40+2v2
50-40=2v2
10=2v2
divide both sides by 2
v2=10/2
v2=5m/s
The pin's final velocity is 5m/s
Answer:
<h2>C. <u>
0.55 m/s towards the right</u></h2>
Explanation:
Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Momentum = Mass (M) * Velocity(V)
BEFORE COLLISION
Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s
Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s
AFTER COLLISION
Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s
<u>x is the final velocity of the 0.25kg ball</u>
Momentum of 0.15kg body moving at 0.75m/s(body at rest) =
0.15 * 0.75kgm/s = 0.1125 kgm/s
Using the law of conservation of momentum;
0.25+0 = 0.25x + 0.1125
0.25x = 0.25-0.1125
0.25x = 0.1375
x = 0.1375/0.25
x = 0.55m/s
Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>
<u></u>
