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kondor19780726 [428]
3 years ago
6

49. A vertically hung 0.50-meter- long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight a

ttached to the spring. If 15 joules of elastic potential energy are stored in the spring , what is the value of the spring constant?
Physics
1 answer:
Natali5045456 [20]3 years ago
4 0

Answer:

K=120

Explanation:

From the question we are told that

Length of spring   l_1=0.5m

Length of stretched l_s=1m

Potential energy of spring E=15J

Generally equation for energy stored is mathematically given as

U=1/2K \triangle x^2

K=\frac{2U}{\triangle x^2}

K=\frac{2*15}{ 0.5^2}

Therefore value of the spring constant in N/m? is given as

K=120

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Explanation:

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C.The north pole of a bar magnet will attract the south pole of another bar magnet.

D.The south poles of two bar magnets will repel each other.

F.Earth's geographic north pole is actually a magnetic south pole.

Explanation:

The magnetic field is a region of space that exerts a force on charged particles in motion or on magnets.

A magnet is an object that produces a magnetic field in the region around it - a magnet always consists of a north pole and of a south pole; it is not possible to create a magnetic monopole: if a magnet is cut in a half, we end up with two new magnets, each of them having its own north and south pole.

The lines showing the direction of the magnetic field always go from the north pole towards the south pole.

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C.The north pole of a bar magnet will attract the south pole of another bar magnet.

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