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sergejj [24]
3 years ago
8

Two charged particles are placed 2.0 meters apart. The first charge is +2.0 E-6 C, and the second charge is +4.0 E-6 C. What is

the electrical force between the two charges?
(k = 9.0 E9 Nm^2/c^2)
A. +1.8 E-2 N and it is repulsive
B. +3.6 E-2 N and it is repulsive
C. +4.5 E-2 N and it is attractive
D. +9.0 E -2 N and it is attractive
Physics
1 answer:
ipn [44]3 years ago
4 0
F = kq1q2/r^2 

<span>q1 is first charge </span>
<span>q2 is second charge </span>
<span>k is 9 E9 </span>
<span>r is distance between them </span>

<span>F = (9E9)(2 E-6)(4 E-6)/2^2 = 0.018 N </span>

<span>A postive answer indicates a repulsive force</span>
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<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then: 

F = G(Mm / (4(pi)*r^2)) 

The above expression gives the force that you feel on the earth's surface, as it is today! 

Let us now double the mass of the earth and decrease its diameter to half its original size. 

This is the same as replacing M with 2M and r with r/2. 

Now the gravitational force (F' ) on the new earth's surface is given by: 

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3 years ago
Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s
g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

v_{x} =  v cos(∅)

v_{y} =  v sin(∅)

We know that for a projectile motion,

t =\frac{2vsin(∅)}{g}

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = v_{x}×t = vcos(∅)×\frac{2vsin(∅)}{g} = \frac{v^{2}sin(2∅) }{g}.

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = \frac{v^{2}sin(2α) }{g} = \frac{v^{2}sin(2[90-β]) }{g} =\frac{v^{2}sin(180-2β) }{g} = \frac{v^{2}sin(2β) }{g} .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

     β = 90-25 = 65°

∴ The required angle is 65°.

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