Answer:
f something happens to go wrong at a nuclear reactor, anyone living in a 10-mile radius of the plant may have to evacuate. This map also shows a 50-mile evacuation zone, the safe distance that the U.S. government recommended to Americans who were near
because
The seasons affect the scheduling of far more events the moon phases do.
Complete Question:
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;

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:

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

Total potential due to this charges = 4500 V + 6750 V = 11250 V
Answer:
2.572 m/s²
Explanation:
Convert the given initial velocity and final velocity rates to m/s:
- 65 km/h → 18.0556 m/s
- 35 km/h → 9.72222 m/s
The motorboat's displacement is 45 m during this time.
We are trying to find the acceleration of the boat.
We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.
Substitute the known values into the equation.
- (9.72222)² = (18.0556)² + 2a(45)
- 94.52156173 = 326.0046914 + 90a
- -231.4831296 = 90a
- a = -2.572
The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².
The answer to this question is true I believe