The star with apparent magnitude 2 is more brighter than 7.
To find the answer, we have to know about apparent magnitude.
<h3>What is apparent magnitude?</h3>
- 100 times as luminous as a star with an apparent brightness of 7 is a star with a magnitude of 2.
- The apparent magnitude of bigger stars is always smaller.
- The brightest star in the night sky is Sirius.
- The brightness of a star or other celestial object perceived from Earth is measured in apparent magnitude (m).
- The apparent magnitude of an object is determined by its inherent luminosity, its distance from Earth, and any light extinction brought on by interstellar dust in the path of the observer's line of sight.
Thus, we can conclude that, the star with apparent magnitude 2 is more brighter than 7.
Learn more about the apparent magnitude here:
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Explanation:
Let the east direction to be i^ and north direction be j^.
Thus displacement of the man S=8i^+6j^
So, magnitude of displacement ∣S∣=62+82=100=10 m
Answer:
The time it takes the ball to stop is 0.021 s.
Explanation:
Given;
mass of the softball, m = 720 g = 0.72 kg
velocity of the ball, v = 15.0 m/s
applied force, F = 520 N
Apply Newton's second law of motion, to determine the time it takes the ball to stop;

Therefore, the time it takes the ball to stop is 0.021 s.
Answer:
10 m/s^2
Explanation:
Equation: F = ma.
a = acceleration
m = mass
F = force
Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.
F = 20
m = 2
a = ?
a = F/m
a = 20/2
a = 10 m/s^2
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - (
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V