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Sergeeva-Olga [200]
3 years ago
9

Now consider the statements in Part A that are inferred from models. A solar model is used to calculate interior conditions base

d on certain "known" characteristics of the Sun, such the Sun’s total mass. How do we know the Sun’s mass?a. We infer the mass from a model of the Sun.b. We first measure the Sun's size and density, then use these measurements to calculate the Sun’s mass.c. We can calculate it by using the law of conservation of energy with the measured amount of light that the Sun emits.d. We can calculate it by applying Newton's version of Kepler's third law with Earth's orbital period (1 year) and Earth’s average distance from the Sun (1 AU).
Physics
1 answer:
mafiozo [28]3 years ago
6 0

Answer:

d. We can calculate it by applying Newton's version of Kepler's third law

Explanation:

The measurements of a Star like the Sun have several problems, the first one is distance, but the most important is the temperature since as we get closer all the instruments will melt. This is why all measurements must be indirect because of the effects that these variables create on nearby bodies.

Kepler's laws are deduced from Newton's law of universal gravitation, in these laws the mass of the Sun affects the orbit of the planets since it creates a force of attraction, if measured the orbit and the time it takes to travel it we can know the centripetal acceleration and with it knows the force, from where we clear the mass of the son.

Let's review the statements of the exercise

.a) False. We don't have good enough models for this calculation

.b) False. The size of the sun is very difficult to measure because it is a mass of gas, in addition the density changes strongly with depth

.c) False. The amount of light that comes out of the sun is not all the light produced and is due to quantum effects where the mass of the sun is not taken into account

.d) True. This method has been used to calculate the mass of the sun and the other planets since the variable distance and time are easily measured from Earth

Correct answer is D

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Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers
Dmitrij [34]

Answer:

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

6 0
3 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

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3 years ago
A substance that contains only one kind of matter is known as a(n)
Anna35 [415]
An element would be the type of substance that only contains one kind of matter.
3 0
3 years ago
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4 0
3 years ago
Two plane mirrors are inclined at an angle of 70. For a ray of light, incident on mirror 1 , its angle of reflection from mirror
tester [92]

Answer:

D

Explanation:

Two plane mirrors are inclined at 70∘. A ray incident on one mirror at incidence angle θ after reflection falls on the second mirror and is reflected from there parallel to the first mirror, The value of θ is. ∴(θ)=50∘.

8 0
3 years ago
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