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NARA [144]
3 years ago
11

Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to

tal electric force that these charges exert on a third point charge Q = μC at x = m, y = 0?
Physics
1 answer:
pishuonlain [190]3 years ago
6 0

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

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The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t
Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

PE=\frac{GMm}{r}

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

m = \rho V

m = (1030Kg/m^3)(7*10^8m^3)

m = 7.210*10^{11}Kg

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m

PART A) Potential energy when the ocean is at its furthest point to the moon,

PE_1 = \frac{GMm}{r_1}

PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}

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PART B) Potential energy when the ocean is at its closest point to the moon

PE_2 = \frac{GMm}{r_2}

PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}

PE_2 = 9.361*10^{15}J

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

\Delta KE = \Delta PE

\frac{1}{2}mv^2 = PE_2-PE_1

v=\sqrt{2(PE_2-PE_1)/m}

v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}

v = 29.4m/s

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