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Tomtit [17]
3 years ago
11

Give a claim about what is affecting the orbital speed and how it affects it

Physics
1 answer:
scoray [572]3 years ago
7 0

Answer:

A planet's orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun's gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun's gravitational pull, and the slower it moves in its orbit

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what is the average gravitational force of attraction between the earth and the sun? the earth averages a distance of about 150
Tanzania [10]

Answer:

B

Explanation:

Hhhhh

5 0
3 years ago
Physics double pivot question​
andriy [413]

Explanation:

Assuming the wall is frictionless, there are four forces acting on the ladder.

Weight pulling down at the center of the ladder (mg).

Reaction force pushing to the left at the wall (Rw).

Reaction force pushing up at the foot of the ladder (Rf).

Friction force pushing to the right at the foot of the ladder (Ff).

(a) Calculate the reaction force at the wall.

Take the sum of the moments about the foot of the ladder.

∑τ = Iα

Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0

Rw (3.0 sin 60°) = mg (1.5 cos 60°)

Rw = mg / (2 tan 60°)

Rw = (10 kg) (9.8 m/s²) / (2√3)

Rw = 28 N

(b) State the friction at the foot of the ladder.

Take the sum of the forces in the x direction.

∑F = ma

Ff − Rw = 0

Ff = Rw

Ff = 28 N

(c) State the reaction at the foot of the ladder.

Take the sum of the forces in the y direction.

∑F = ma

Rf − mg = 0

Rf = mg

Rf = 98 N

3 0
3 years ago
I NEED THIS ASAP!!! A ball is thrown straight up with an initial velocity of 4.40 m/s. Assuming there is no air friction, what i
Lilit [14]

Answer:

I think it is 80m/s

Explanation:

d = ½ g t2

  = ½ (10 m/s2) (4 s)2

  = (5 m/s2) (16 s2)

  = 80 ms

75% sure

Hope this helps!!!

5 0
3 years ago
A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box
Zielflug [23.3K]

m = mass of the box

N = normal force on the box

f = kinetic frictional force on the box

a = acceleration of the box

μ = coefficient of kinetic friction

perpendicular to incline , force equation is given as

N = mg Cos30                                         eq-1

kinetic frictional force is given as

f = μ N

using eq-1

f = μ mg Cos30    


parallel to incline , force equation is given as

mg Sin30 - f = ma

mg Sin30 - μ mg Cos30  = ma

"m" cancel out

a = g Sin30 - μ g Cos30

inserting the values

1.20 = (9.8) Sin30 - (9.8) Cos30 μ

μ = 0.44

4 0
3 years ago
Your mom is making lunch which forces are acting on your mom
N76 [4]

The forces acting on your mom while cooking is Air resistance and the force of friction

<u>Explanation:</u>

<u>1. Air resistance:</u>

  • In simple words, Air resistance can be stated as the type of friction  between the air and the other materials.
  • In this scenario, there will be an air resistance and the air hits the mom while cooking via the doors or windows

<u>2. The force of friction:</u>

  • In simple words, friction can be stated as, the resistance that one surface or object encounters when moving over another.
  • While cooking the food mom would experience the friction since friction is the transfer of heat, and cooking is the process of receiving that heat.

7 0
3 years ago
Read 2 more answers
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