A star's motion towards or away from us is detected by the ________ of the spectral lines.

1 answer:

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<span>A star's motion towards or away from us is detected by the wavelength of the spectral lines.
Because when a star is moving toward us or away from us, the apparent color does not change but there is a change in the wavelength of spectral lines, when the stars are moving towards us, the spectral lines are towards red end and when they are moving away from us the spectral lines are towards the blue end.</span>

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Suppose you throw a baseball downward from a roof so that it initially has 120 J of gravitational potential energy, and 10 J of

Volgvan

Answer:

It will be greater than 10 J.

Explanation:

The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):

E = PE + KE

According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.

The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.

The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.

Here at the beginning, when it is at the top of the roof, the baseball has:

PE = 120 J

KE = 10 J

So the total energy is

E = 120 + 10 = 130 J

As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).

Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.

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3 years ago

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A 0.311 kg tennis racket moving 30.3 m/s east makes an elastic collision with a 0.0570 kg ball moving 19.2 m/s west. Find the ve

Troyanec [42]

The velocity of tennis racket after collision is 14.96m/s

<u>Explanation:</u>

Given-

Mass, m = 0.311kg

u1 = 30.3m/s

m2 = 0.057kg

u2 = 19.2m/s

Since m2 is moving in opposite direction, u2 = -19.2m/s

Velocity of m1 after collision = ?

Let the velocity of m1 after collision be v

After collision the momentum is conserved.

Therefore,

m1u1 - m2u2 = m1v1 + m2v2

$v1 = (\frac{m1-m2}{m1+m2})u1 + (\frac{2m2}{m1+m2})u2$

$v1 = (\frac{0.311-0.057}{0.311+0.057})30.3 + (\frac{2 X 0.057}{0.311 + 0.057}) X-19.2\\\\v1 = (\frac{0.254}{0.368} )30.3 + (\frac{0.114}{0.368}) X -19.2\\ \\v1 = 20.91 - 5.95\\\\v1 = 14.96$