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mihalych1998 [28]
3 years ago
12

Kalyan ramji sain, of india, had a mustache that measured 3.39 m from end to end in 1993. suppose two charges, q and 3q, are pla

ced 3.39 m apart. if the magnitude of the electric force between the charges is 2.4 × 10−6 n, what is the value of q?
Physics
1 answer:
jeka943 years ago
4 0

Answer:

q=3.19*10^{-8}C

Explanation:

The expression for the electric force between two charges

F=k\frac{q_1q_2}{r^2}\\

where K is the Coulomb's constant (k=9*10^{9}Nm^2/C^2), and we have that

q1=q

q2=3q

r=3.39m

F=2.4*10^{-6}N

By replacing in the formula we have

F=2.4*10^{-6}N=(9*10^{9}\frac{Nm^2}{C^2})\frac{(q)(3q)}{(3.39m)^2}\\\\

and by taking apart q we have

q=3.19*10^{-8}C

hope this helps!!

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A particle moves along line segments from the origin to the points (1, 0, 0), (1, 5, 1), (0, 5, 1), and back to the origin under
Kaylis [27]

Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

4 0
3 years ago
*15 points*
Svetradugi [14.3K]
Same as the other person most likely would be 20 times louder as your answer
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6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.
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Answer:

acceleration of the rocket is given as

a = 7.45 m/s^2

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

v_i = 0

v_f = 447 m/s

t = 1 min = 60 s

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a = \frac{v_f - v_i}{t}

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When the Sun is between the Earth and Neptune, the distance is found by the relation;

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The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

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