The net force on particle particle q1 is 13.06 N towards the left.
<h3>
Force on q1 due to q2</h3>
F(12) = kq₁q₂/r₂
F(12) = (9 x 10⁹ x 13 x 10⁻⁶ x 7.7 x 10⁻⁶)/(0.25²)
F(12) = -14.41 N (towards left)
<h3>Force
on q1 due to q3</h3>
F(13) = (9 x 10⁹ x 7.7 x 10⁻⁶ x 5.9 x 10⁻⁶)/(0.55²)
F(13) = 1.352 N (towards right)
<h3>Net force on q1</h3>
F(net) = 1.352 N - 14.41 N
F(net) = -13.06 N
Thus, the net force on particle particle q1 is 13.06 N towards the left.
Learn more about force here: brainly.com/question/12970081
#SPJ1
Answer:
5235.84 kg
Explanation:
There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.
As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh,
;

Answer:
Explanation:
ASSUMING your speed is constant
f₀ = f(v + vo)/(v + vs)
Δf = f approach - f depart
69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))
69.5 = 769(2vo/343)
vo = 15.5 m/s
Answer:
0.96 m, upward
Explanation:



Initial velocity, u=0


Where 
Substitute the values





Hence,the block moves upward because displacement is positive.
A set of data has a mean of 12 and a standard deviation of 3. A data point of the set has a z-score of 1.3. What does a z-score of 1.3 mean?
The data point is 1.3 standard deviations away from 3
The data point is 1.3 standard deviations away from 12.
The data point is 3 standard deviations away from 1.3.
The data point is 3 standard deviations away from 12.
its B