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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The gravitational potential energy
gpe = mgh
Answer: A capacitor.
Explanation:
The capacitor is a passive element that is used in electronics to store electrical energy maintaining an electrical field. The simpler case of a capacitor is the parallel plates capacitor.
It consists of two parallel metal plates separated by a distance D, in this case, the air between the plates works as a dielectric, as the plates do not touch each other and are separated by a dielectric, the charge is stored in the surface plates.
There are a lot of other types of capacitors, the most used in actuality may be the cylindrical one, where instead of parallel plates, it uses two concentric cylinders, and the space between the cylinders is filled with a dielectric/insulator.
Answer:
Data:-m=0.88kg ,g=9.8m/sec² ,P.E=96J ,h=?
Explanation:
solution ,P.E=mgh here we have to find h so h=P.E/mg ,h=96/0.88×9.8 ,h=96/8.624=11.131m and if you want to verify so just put the value of h in same formula, likewise :-P.E=mgh ,P.E=0.88×9.8×11.131=96J so we got the same value of P.E as it is given the question (verified).
Answer:
to overcome the out of friction we must increase the angle of the plane
Explanation:
To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.
X axis
fr - Wₓ = m a (1)
Y axis
N- 
= 0
N = W_{y}
let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
the friction force has the formula
fr = μ N
fr = μ Wy
fr = μ mg cos θ
from equation 1
at the point where the force equals the maximum friction force
in this case the block is still still so a = 0
F = fr
F = (μ mg) cos θ
We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.
This is the force that balances the friction force, any force slightly greater than F initiates the movement.
Consequently, to overcome the out of friction we must increase the angle of the plane
the correct answer is to increase the angle of the plane
