2. What is the wavelength of the wave shown above? a. 6m b. 11 C 2m d. 3m
1 answer:
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Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.
From the definition of apparent magnitude, we know that:
where:
m = apparent magnitude
F = corresponding flux
We also know that the flux is given by the formula:
where:
L = luminosity
d = distance
Therefore:
Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
Now, let's solve for m:
= <span>
</span>
= 13
Hence, the apparent magnitude of the star would be m = +13
where:
m = apparent magnitude
F = corresponding flux
We also know that the flux is given by the formula:
where:
L = luminosity
d = distance
Therefore:
Now, let's apply these formulae to the same star (therefore, same luminosity), using apparent magnitude and absolute magnitude (which is defined as the apparent magnitude the star would have if it were at a distance of 10pc):
Now, let's solve for m:
= <span>
= 13
Hence, the apparent magnitude of the star would be m = +13