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Delicious77 [7]
3 years ago
6

Please help i need help i’ll give lots of points

Physics
2 answers:
N76 [4]3 years ago
8 0

Answer:

OK so ik this but what is you question?

Explanation:

erma4kov [3.2K]3 years ago
3 0

Answer:

whats the question first

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OXYGEN

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the mass of a lump of gold is constant everywhere but its weight isnt explain both in weight and mass
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Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p
Alexxandr [17]

Answer:

magnitude of A − B =  15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a  magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = \sqrt{(-15)^{2}+(-5)^{2}}

|A - B| = \sqrt{250}

|A - B| = 15.81 km

7 0
3 years ago
(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.
LiRa [457]

Answer:

a)  [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

             V = k q / r

             V = [N m² / C²] C / m

              V = [N m / C]

The electric potential is defined as

             V = E .s

             V = [N / C] [m]

             V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

            [volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

             v =  - \frac{dE}{dx} i^

therefore the equipotential lines must also be perpendicular to the electric field lines

4 0
3 years ago
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