Select the correct answer.
In general, how does an increase in distance from the Sun affect a planet?
ОА. .
The amount of sunlight the planet receives decreases.
OB.
The gravitational force exerted by the Sun increases.
ОС.
The thickness of a planet's atmosphere increases.
OD. The planet's atmosphere absorbs more heat.

Answer:

Option 2

Explanation:

As per the relation between the distance of the galaxy and shifting of the light of the galaxy towards any specific wavelength of the electromagnetic spectrum, a galaxy at great distance shifts more towards the red spectra that has the highest wavelength.

Thus, this observation give details about the distance of the galaxy from earth.

Answer:

b

Explanation:

Answer:

#1 Yes

Explanation: #1: The rest of them are used mainley by farmers, and crops are used by common citizens in the world.

Question 1: Crops.

Question 2: Diagnostic Services.

Question 3: A cable company needs to lay new fiber optic cable to reach its customers across a large lake.

Question 4: A bachelor's degree in energy research.

Question 5: Environmental Resources.

If any of these answers are incorrect, please tell me, so I can fix my mistake. Thank you.

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²