Answer:
stars will emit more light due to their Luminosity, so they look very bright.
Explanation:
Luminous refers to..,
- The total amount of energy radiated by a star or other celestial object per second.
- Therefore it is the power output of a star.
Most of the really bright stars in our sky are not that very close to us yet they look bright because of the Luminosity of the star.
These stars are intrinsically so luminous.
A star's power output across all wavelengths is called its bolometric luminosity.
A star with large luminosity will have more measure of radiated electromagnetic power meaning.
so it will emit more light than a low luminosity star.
Hence,
those stars can easily be seen even across great distance.
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true if you are refering to the desing of the experimnt as it does identify the variable
Answer:
Mass of the climber = 69.38 kg
Explanation:
Change in length

Load, P = m x 9.81 = 9.81m
Young's modulus, Y = 0.37 x 10¹⁰ N/m²
Area

Length, L = 15 m
ΔL = 5.1 cm = 0.051 m
Substituting
Mass of the climber = 69.38 kg
The floor exerts 20 N of force on the chair
Explanation:
We can answer this question by using Newton's third law, which states that:
<em>"When an object A exerts a force (called action) on an object B, object B exerts an equal and opposite force (called reaction) on object A"</em>
In this problem, we can identify:
- Object A as the chair
- Object B as the floor
This means that the force of 20 N exerted by the chair on the floor is the action, and so the force exerted by the floor on the chair is the reaction. Newton's third law states that these two forces are equal and opposite: therefore, the force exerted by the floor on the chair is also 20 N, but in the opposite direction.
Learn more about Newton's third law:
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The formula we can use in this case is:
d = v0t + 0.5 at^2
v = at + v0
where,
d = distance travelled
v0 = initial velocity = 0 since at rest
t = time travelled
a = acceleration
v = final velocity when it took off
a. d = 0 + 0.5 * 3 * 30^2
d = 1350 m
b. v = 3 * 30 + 0
<span>v = 90 m/s</span>