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erastova [34]
2 years ago
7

The fundamental resolution of a lens or telescope is determined by:

Physics
1 answer:
Sliva [168]2 years ago
4 0

Answer:

The numerical aperture of the lens.

Explanation:

A numerical aperture is a dimensionless number that is the range of angles that the system can accept / emit light.

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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How much meters is a mile
elena-s [515]

Roughly 1609 meters in one mile

6 0
3 years ago
People living at Earth’s equator are traveling at a constant speed of about 1,670 km/h as Earth spins on its axis. What are thes
Vlada [557]
The people living around the equator experience acceleration. Acceleration is the change in speed or/and direction. When on a rotating body, the speed does not change it is only the direction that changes as it rotates.When on a rotating body you do not have to change speed but only direction. In a normal situation when you are traveling on a circle you will never have a straight line.
4 0
3 years ago
A piece of paper looks white in both the noonday sun and under moonlight, even though there is less light being reflected off th
ddd [48]
Answer: Brightness consistency

There are three types of perceptual consistency
Types of Perceptual Constancy: Shape, Size, and Brightness Size constancy
Since the moon and sun affect light, brightness consistency is occurring.
Brightness constancy is our visual ability to perceive objects as having the same level of brightness even though the level of lighting changes.
7 0
3 years ago
What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
Un ciclista en una cuesta hacia abajo, marcha a una velocidad de 45km/h aplica los frenos y al cabo de 5s su velocidad se ha red
nexus9112 [7]

Answer:

Los 0.0416km

esto se debe a que transponemos la fórmula acelerada y obtenemos Distancia = velocidad × tiempo

también recuerda transponer los segundos a horas viendo que la velocidad es por hora

También tenga en cuenta que no hablo español, así que esto fue extremadamente difícil

culto

8 0
2 years ago
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