Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:
You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:
hence, the speed of the Hubble is approximately 384km/min
Answer:
<u></u>
- <u>1. The potential energy of the swing is the greatest at the position B.</u>
- <u>2. As the swing moves from point B to point A, the kinetic energy is increasing.</u>
Explanation:
Even though the syntax of the text is not completely clear, likely because it accompanies a drawing that is not included, it results clear that the posittion A is where the seat is at the lowest position, and the position B is upper.
The gravitational <em>potential energy </em>is directly proportional to the height of the objects with respect to some reference altitude. Thus, when the seat is at the position A the swing has the smallest potential energy and when the seat is at the <em>position B the swing has the greatest potential energy.</em>
Regarding the forms of energy, as the swing moves from point B to point A, it is going downward, gaining kinetic energy (speed) at the expense of the potential energy (losing altitude). When the seat passes by the position A, the kinetic energy is maximum and the potential energy is miminum. Then the seat starts to gain altitude again, losing the kinetic energy and gaining potential energy, up to it gets to the other end,
Answer:
The answers to the questions have been solved in the attachment.
Explanation:
The answers to part a to e are all contained in the attachment. For answer part b, temperature and frequency were assumed to be fixed or constant. V² is directly proportional to T telling us that variation in T gives us a square in the frequency variation. This tells us why it is difficult when both frequencies are on this side of the black body.
The number of orders that one can see the entire visible speed is 5 orders.
<h3>How to calculate the orders?</h3>
From the information given, it should be noted that the grating spacing will be:
d = (1.00 × 10^-4) / 250
d = 4000nm
Therefore, the number of times that are needed to complete the order will be the same as the number of orders which the long wavelength time will be visible. This will be:
= (4000 × sin 90°)/700
= 5.71
Therefore, the maximum orders will be 5.
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