Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)

Answer:

The potential due to these charges is 11250 V

Explanation:

Potential V is given as;

$V =\frac{Kq}{r}$

where;

K is coulomb's constant = 9x10⁹ N.m²/C²

r is the distance of the charge

q is the magnitude of the charge

The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{6} =4500 V$

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

$V =\frac{9X10^9 X3X10^{-6}}{4} =6750 V$

Total potential due to this charges = 4500 V + 6750 V = 11250 V

3 0

4 years ago

already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you

Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0

3 years ago

A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.9m/s and the hatchback ca

4vir4ik [10]

Answer:

s = 589.3 m

Explanation:

Let the truck and car meet at a distance = s m

The truck is moving at constant velocity = v

so s= v * t ---------- (1)

car:

Vi = 0 m/s

a = 3.9 m/s²

s = Vi* t + 1/2 a t²

s= 0 * t + 1/2 a t²

s = 1/2 a t² ----------- (2)

compare equation (1) and equation (2)

s= v * t = 1/2 a t²

⇒ v * t = 1/2 a t²

⇒ t = 2 * v/ a

⇒ t = (2 * 33.9 )/ 3.9

⇒ t = 17. 38 s

Now

from equation (1)

s= v * t

s= 33.9 * 17.38

⇒ s = 589.3 m

3 0

3 years ago