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
You would be correct.
Because you have only JUST released the arrow, and how close he is to the target, it would have the same amount of energy when it strikes the target. Yes, the kinetic energy would be destroyed when you hit the target but not right away. And yes, the potential energy would also be destroyed once you release the arrow, but it goes straight back once it stops moving, aka when it hits the target, although it has only just stopped moving.
Hope this helps!
An atom. Hope this helped
Answer:
Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
<u><em>Now,</em></u>
From the question,
V(t) = 230 sin (100t) ---------------(ii)
<em><u>By comparing equations (i) and (ii) the following holds;</u></em>
V = 230
ω = 100
Ф = 0
<em><u>But;</u></em>
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f =
Hz
Therefore, the frequency of the voltage is
Hz