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elena-14-01-66 [18.8K]
3 years ago
10

What is the difference between a universal law and a scientific theory?

Physics
1 answer:
marissa [1.9K]3 years ago
7 0

C)


Scientific theories can change with new evidence, but Universal Laws are unchanging and always in effect everywhere.

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How do you determine the wattage capacity needed by a power supply?
Lady bird [3.3K]
We could use the formula for the Power supply in order to find the wattage capacity and it would be:

P = V²/R or P = V * I

Hope this helps!
6 0
2 years ago
Two hypothetical discoveries in Part A deal with moons that, like Earth's moon, are relatively large compared to their planets.
ser-zykov [4K]

Answer:

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible.

6 0
3 years ago
What do the spheres in this model represent?
san4es73 [151]
I believe it’s B. Electrons
7 0
3 years ago
Dinah is playing cards with her friends, and it is her turn to deal. A card with a mass of 2.3 g slides 0.35 m along the table b
Inga [223]

Answer:

soln,

=2.3+0.35-0.24

=2.41 answer

4 0
2 years ago
Two 3.0 μC charges lie on the x-axis, one at the origin and the other at What is the potential (relative to infinity) due to the
Airida [17]

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;

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

6 0
3 years ago
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