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.
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Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;

Therefore, the frequency of this mode of vibration is 138.87 Hz
Answer:
70.5 mph
Explanation:
A passenger jet travels from Los Angeles to Bombay, India, in 22h.
The return flight takes 17 h.
The difference in flight times is caused by winds over the Pacific Ocean that
blow primarily from west to east.
If the jet's average speed in still air is 550 mi/h what is the average speed
of the wind during the round trip flight? Round to the nearest mile per hour.
Is your answer reasonable?
:
Let w = speed of the wind
:
Write a distance equation (dist is the same both ways
17(550+w) = 22(550-w)
9350 + 17w = 12100 - 22w
17w + 22w = 12100 - 9350
39w = 2750
W = 2750/39
w = 70.5 mph seems very reasonable
:
Confirming if the solution by finding the distances using these value
17(550+70.5) = 10549 mi
22(550-70.5) = 10549 mi; confirms our solution of w = 70.5 mph
Formula: s = d/t
s = speed
d = distance
t = time
Solve using the values we are given.
s = 300/40
s = 7.5m/s
Best of Luck!
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