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andreyandreev [35.5K]
3 years ago
13

Assignment 10 Coulombic Equation Practice Directions: Complete the following problems to calculate the electrostatic force that

is present in each situation. You will need to be sure to correctly substitute each of the variables in the equation. Once again the Equation is; F=K[(q1*q2)/r2]. Also Coulomb’s Constant is 8.98*109. What is the magnitude of the force a 1.5 x 106C charge exerts on a 3.2 x 104C charge located 1.5 m away?
Physics
1 answer:
Tatiana [17]3 years ago
8 0

Answer:

Magnitude of the force between the charges is F = 1.92×10^20N

Explanation:

Given the magnitude of force according to coulombs law

F =K[(q1*q2)/r2]

Where q1 and q2 are the charges

r is the distance between the charges

K is the coulombs constant

Substituting the given values, we have;

F = 8.98×10^9 × 1.5×10^6 × 3.2×10^4/1.5²

F = 43.1×10^19/2.25

F = 19.16×10^19N

F = 1.92×10^20N

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A large truck is moving 22.0 m/s. If it’s momentum is 125,000 kg • m/s, what is the trucks mass
strojnjashka [21]

Answer:

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Explanation:

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steps:

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Inez uses hairspray on her hair each morning before going to school. The spray spreads out before reaching her hair partly becau
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Answer:

7.0\cdot 10^{-13}C

Explanation:

The magnitude of the electrostatic force between two charged objects is

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the two charges

r is the separation between the two charges

The force is attractive if the charges have opposite sign and repulsive if the charges have same sign.

In this problem, we have:

r=0.070 cm =7\cdot 10^{-4} m is the distance between the charges

q_1=q_2=q since the charges are identical

F=9.0\cdot 10^{-9}N is the force between the charges

Re-arranging the equation and solving for q, we find the charge on each drop:

F=\frac{kq^2}{r^2}\\q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(9.0\cdot 10^{-9})(7\cdot 10^{-4})^2}{8.99\cdot 10^9}}=7.0\cdot 10^{-13}C

8 0
3 years ago
If the earth's magnetic field has strength 0.50 gauss and makes an angle of 20.0 degrees with the garage floor, calculate the ch
lys-0071 [83]

Answer:

ΔΦ = -3.39*10^-6

Explanation:

Given:-

- The given magnetic field strength B = 0.50 gauss

- The angle between earth magnetic field and garage floor ∅ = 20 °

- The loop is rotated by 90 degree.

- The radius of the coil r = 19 cm

Find:

calculate the change in the magnetic flux δφb, in wb, through one of the loops of the coil during the rotation.

Solution:

- The change on flux ΔΦ occurs due to change in angle θ of earth's magnetic field B and the normal to circular coil.

- The strength of magnetic field B and the are of the loop A remains constant. So we have:

                         Φ = B*A*cos(θ)

                         ΔΦ = B*A*( cos(θ_1) - cos(θ_2) )

- The initial angle θ_1 between the normal to the coil and B was:

                         θ_1  = 90° -  ∅

                         θ_1  = 90° -  20° = 70°

The angle θ_2 after rotation between the normal to the coil and B was:

                         θ_2  =  ∅

                         θ_2  = 20°

- Hence, the change in flux can be calculated:

                        ΔΦ = 0.5*10^-4*π*0.19*( cos(70) - cos(20) )

                        ΔΦ = -3.39*10^-6

                       

8 0
3 years ago
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