Answer:
The beam of light is moving at the peed of:
km/min
Given:
Distance from the isalnd, d = 3 km
No. of revolutions per minute, n = 4
Solution:
Angular velocity, 
(1)
Now, in the right angle in the given fig.:
Now, differentiating both the sides w.r.t t:
Applying chain rule:
Now, using 
and y = 1 in the above eqn, we get:
Also, using eqn (1),
The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.
Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima. The latitude in the middle
of that intersection is 46.585° North. <u>That's</u> the number we need.
Here's how I would do it:
-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.
-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°.
-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.
This sets the limit of the highest in the sky that the moon can ever appear.
90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .
That doesn't happen regularly. It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).
Depending on the time of year, that can be any time of the day or night.
The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.
In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky. Then it's going to be somewhere near
67° above the horizon at midnight.
Gravitational forces are forces of attraction. It's like the Earth pulling on you and keeping you on the ground.
Answer:
A.) 27000 kgm/s
18000 kgm/s
B.) Va = 22 m/s
C.) 19800 kgm/s
25200 kgm/s
Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g
M = 9×10^5/1000 = 900 kg
A.) Initial momentum of A
Mu = 900 × 30 = 27000 kgm/s
Initial momentum of B
Mu = 900 × 20 = 18000 kgm/s
B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.
Momentum before impact = momentum after impact
Given that Vb = 28 m/s
27000 + 18000 = 900Va + 900 × 28
45000 = 900Va + 25200
900Va = 45000 - 25200
900Va = 19800
Va = 19800/900
Va = 22 m/s
C.) Momentum of A after impact
MV = 900 × 22 = 19800 kgm/s
Momentum of B after impact
MV = 900 × 28 = 25200 kgm/s
Answer:
undergoes a transition to a quantum state of lower energy
Explanation:
When electrons in an atom move to another quantum state, they emit/absorb a photon according to the following:
- If the electron is moving to a higher energy state, it absorbs a photon (because it needs energy to move to a higher energy level, so it must absorb the energy of the photon)
- if the electron is moving to a lower energy state, it emits a photon (because it releases the excess energy)
In particular, the energy of the absorbed/emitted photon is exactly equal to the difference in energy between the two levels of the electron transition:
