For this problem the figure below shows the representation of a student who pulls on a 20kg box. We know this variables:
Weight of the box = 20kg
Force used by the student to pull on the box = 50N (This is the tension T)
Angle relative to the horizontal = 45 degrees
Aceleration of the box =
The figure also shows the Free-Body diagram, Applying Newton's Second Law we can find the equation for this diagram, related to the x-axis as:
Isolating
:
<span>That is the friction force on the box.</span>
Fe=K Q1/Q2/d2
Q1 is the first charge
Q2 is the second charge
d is the distance
K= 9x10^9 NM^2/C2
Now let’s plug the numbers
Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges
Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment
And we can curtailment the M^2and the other M^2
Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3
Let’s multiply the (9)(2)(8)=144
And add the exponents (9)+(-4)+(-4)=1
So now we got 144x10N divide by the distance which is 0.3
144x10N / 0.3 = 4800N
Hope it helps u understand :)
The correct answer is (B)
Which is (kQ1Q2) / d^2
Never mind i’m pretty sure
A geostationary orbit can be achieved only at an altitude very close to 35,786 km (22,236 mi) and directly above the equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) and an orbital period of 1,436 minutes, which equates to almost exactly one sidereal day (23.934461223 hours).