Answer:
And the force of ( Attraction or repulsion) between the poles A and D ( maximum or minimum)
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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To calculate force, use the formula force equals mass times acceleration, or F = m × a. Make sure that the mass measurement you're using is in kilograms and the acceleration is in meters over seconds squared. When you've solved the equation, the force will be measured in Newtons.
Answer:
Explanation:
Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg
Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²
For change in period of rotation we shall apply conservation of angular momentum law
I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .
I₁ / I₂ = ω₂ / ω₁
I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .
T₂ / T₁ = I₂ / I₁
(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²
1 + 5 / 3 x 2.3 x 10¹⁹ / M
= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴
= 1 + .0000064
T₂ = 24 (1 + .0000064)
= 24 hours + .55 s
change in length of the day = .55 s .
Answer:
230 m/s northeast, 1.8 m/s up
Explanation:
204 kilometres = 204000 metres
15.0 minutes = 900 seconds
Velocity = Distance / Time
= 204000 / 900
= 230 m/s northeast (to 2 sf.)
1.6km = 1600 metres
Velocity = 1600 / 900
= 1.8 m/s up (to 2 sf.)
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