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alexandr1967 [171]
3 years ago
15

Why is polaris used as a celestial reference point?

Physics
1 answer:
ser-zykov [4K]3 years ago
8 0

At the equator, the North Celestial and South Celestial Poles would lie on the horizon where the meridian intersects the horizon. Polaris is called the "North Star." ... Polaris is special because the Earth's North Pole points almost exactly towards it.

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Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of 250 N to the left, and Janet pulls with a fo
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Answer: 75 N to the right

Explanation:

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What is the formula for conservation of momentum
olga55 [171]

Answer:

The general equation for conservation of momentum during a collision between n number of objects is given as: [m i ×v i a ] = [m i ×v i b ] Where m i is the mass of object i , v i a is the velocity of object i before the collision, and v i b is the velocity of object i after the collision.

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Regions in the milky way where density waves have caused gas clouds to crash into each other are called
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3 years ago
A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)
Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

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wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0
2 years ago
a mass on a spring vibrates in simple harmonic motion at an amplitude of 8.0 cm. if the mass of the object is 0.20kg and the spr
Reil [10]

Answer:

4.06 Hz

Explanation:

For simple harmonic motion, frequency is given by

f=\frac {1}{2\pi}\times \sqrt{\frac {k}{m}} where k is spring constant and m is the mass of the object.

Substituting 0.2 Kg for mass and 130 N/m for k then

f=\frac {1}{2\pi}\times \sqrt{\frac {130}{0.2}}=4.057670803\\f\approx 4.06 Hz

5 0
3 years ago
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