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3 years ago

Unexpected orbital velocities of stars around the centers of galaxies led astronomers to predict the existence of dark _____.

Physics

2 answers:

velikii [3]3 years ago

8 0

Answer:

matter

Explanation:

The stars are orbiting around a matter which is not visible. This matter is called dark matter. They are very massive thus, their force of attraction is very large.

There is lot of dark matter in our universe.

Molodets [167]3 years ago

6 0

Unexpected orbital velocities of stars around the centers of galaxies led astronomers to predict the existence of dark matter. Dark matters are hypothetical substance that are believed to account for around five-sixths of the matter in the universe.

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A cat lies on the floor. can you say that no force acts on the cat? or is it correct to say that no net force acts on the cat? e

DIA [1.3K]

Odd though it seems at first, gravity is pulling the cat down while the floor is pushing the cat up - in equal amounts. Forces are absolutely acting on the cat but they balance - so there is no net force.

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2 years ago

What is the length of a spring that has 450J of potential energy and a spring constant of 650N/m?

11111nata11111 [884]

Answer:

Δx = 1.2 m

Explanation:

The CHANGE of spring length) (Δx) can be found using PS = ½kΔx²

Δx = √(2PS/k) = √(2(450)/650) = 1.17669... ≈ 1.2 m

The actual length of the spring is unknown as it varies with material type, construction method, extension or compression, and other variables we have no clue about.

4 0

3 years ago

A trampoline spring has a force constant k = 800 N/m and is stretched exactly 17.5cm. What is the energy required to do this?

Artist 52 [7]

Answer:

the energy required for the extension is 12.25 J

Explanation:

Given;

force constant of trampoline spring, k = 800 N/m

extension of trampoline spring, x = 17.5 cm = 0.175 m

The energy required for the extension is calculated as;

E = ¹/₂kx²

E = 0.5 x 800 x 0.175²

E = 12.25 J

Therefore, the energy required for the extension is 12.25 J

6 0

2 years ago

A box is placed on a 30o frictionless incline. What is the acceleration of the box as it slides down the incline

balandron [24]

Answer:

2.78m/s²

Explanation:

Complete question:

A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?

According to Newton's second law of motion:

$\sum F_x = ma_x\\F_m - F_f = ma_x\\mgsin\theta - \mu mg cos\theta = ma_x\\gsin\theta - \mu g cos\theta = a_x\\$

Where:

$\mu$ is the coefficient of friction

g is the acceleration due to gravity

Fm is the moving force acting on the body

Ff is the frictional force

m is the mass of the box

a is the acceleration'

Given

$\theta = 30^0\\\mu = 0.25\\g = 9.8m/s^2$

Required

acceleration of the box

Substitute the given parameters into the resulting expression above:

Recall that:

$gsin\theta - \mu g cos\theta = a_x\\$

9.8sin30 - 0.25(9.8)cos30 = ax

9.8(0.5) - 0.25(9.8)(0.866) = ax

4.9 - 2.1217 = ax

ax = 2.78m/s²

Hence the acceleration of the box as it slides down the incline is 2.78m/s²

5 0

3 years ago

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3 years ago