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adelina 88 [10]
3 years ago
5

Two metal balls have charges of 7.1 × 10-6 coulombs and 6.9 × 10-6 coulombs. They are 5.7 × 10-1 meters apart. What is the force

of interaction between the two balls? (k = 9.0 × 109 newton·meters2/coulomb2)
Physics
2 answers:
Murljashka [212]3 years ago
6 0
The answer is 1.4 newtons <span />
Vanyuwa [196]3 years ago
4 0

Answer : The force of interaction between them is 1.35 N

Explanation :

Charge 1 q_1=7.1\times 10^{-6}\ C

Charge 2,  q_2=6.9\times 10^{-6}\ C  

Distance between charges, d=5.7\times 10^{-1}\ m

The force of interaction between them is given by using Coulomb's law. It is written as :

F=k\dfrac{q_1q_2}{d^2}

k is electrostatic constant

F=9\times 10^9\times \dfrac{7.1\times 10^{-6}\ C\times 6.9\times 10^{-6}\ C}{(5.7\times 10^{-1}\ m)^2}

F=1.35\ N

Hence, this is the required solution.

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1 year ago
A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum c
harkovskaia [24]

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, v_1=2.58 m/s

a.r=75.1 cm=75.1\times 10^{-2}m=0.751 m

1cm=10^{-2} m

Tension in the pendulum cable is given  by

Tension=Centripetal force+force due to gravity

T=\frac{mv^2}{r}+mg

Where g=9.8 m/s^2

Substitute the values

T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8

T=8.45 N

b.When the pendulum reaches its highest point,then

Final velocity, v_2=0

According to law of conservation of energy

mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2

gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2

h_1=0

Substitute the values

9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2

3.3282=9.8h_2

h_2=\frac{3.3282}{9.8}=0.34 m

The angle mad  by cable with the vertical=cos\theta=\frac{0.751-0.34}{0.751}=0.55

\theta=cos^{-1}(0.55)=56.6^{\circ}

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension  in the pendulum cable

T=mgcos\theta

Substitute the values

T=0.453\times 9.8cos56.6

T=2.4 N

8 0
3 years ago
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