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horsena [70]
3 years ago
7

Listed following are three possible models for the long-term expansion (and possible contraction) of the universe in the absence

of dark energy. Rank each model from left to right based on the ratio of its actual mass density to the critical density, from smallest ratio (mass density much smaller than critical density) to largest ratio (mass density much greater than critical density).Smallest to largesta.coasting universeb.critical universec.recollapsing universe
Physics
1 answer:
sineoko [7]3 years ago
7 0

Answer:

From smallest ratio to the largest ratio:

Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)

Explanation:

The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1

The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1

Recollapsing Universe: The expansion of the universe reverses in the future and the universe eventually recollapses. The recollapsing universe has the ratio of the actual density to the critical density to be greater than 1

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A 330 kg piano slides 3.6 m down a 28o incline and is kept from accelerating by a man who is pushing back on it parallel to the
e-lub [12.9K]

Answer:

a. 652.68N

b. -2349.65J

c. -3116.12J

d. 5465.77J

e. Zero

Explanation:

a. According to equilibrium of forces, the force of gravity is equal to the sum of the frictional force and force exerted by the man in the opposite direction (since they're both resistant forces).

Fg = Fm + Fr

Fm = Fg - Fr

Fm = mgsin(28°) - umgcos(28°)

u = coefficient of frictional force.

Fm = 330*9.8*sin28 - 0.4*330*9.8*cos28

Fm = 1518.27 - 865.59

Fm = 652.68N

b. Work done by man is:

Wm = -Fm * d

Wm = -652.68 * 3.6

Wm = -2349.65J

c. Work done by friction force:

W(Fr) = -Fr * d

W(Fr) = -865.59 * 3.6

W(Fr) = -3116.12J

d. Work done by gravity:

Wg = Fg * d

Wg = 1518.27 * 3. 6

Wg = 5465.77J

e. Net work done on the piano is:

Work done by friction + work done by gravity + work done by man

= -3116.12 + 5464.77 + (-2349.65)

= 0J

7 0
3 years ago
A cannon tilted upward at 30° fires a cannonball with a speed of 100 m/s. At that instant, what is the component of the cannonba
Sonja [21]

Answer:

the cannonball’s velocity parallel to the ground is 86.6m/S

Explanation:

Hello! To solve this problem remember that in a parabolic movement the horizontal component X of the velocity of the cannonball is constant while the vertical one varies with constant acceleration.

For this case we must draw the velocity triangle and find the component in X(see atached image).

V= Initial velocity=100M/S

cos30=\frac{Vx}{V}

V= Initial velocity=100M/S

Vx=cannonball’s velocity parallel to the ground

Solving for Vx

Vx=Vcos30

Vx=(100m/S)(cos30)=86.6m/s

the cannonball’s velocity parallel to the ground is 86.6m/S

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3 years ago
At what distance above the surface of the earth is the gravitational field 4.9 m/s^2
san4es73 [151]

That's 1/2 of what it is on the surface.

The distance between the center of the Earth and any object
on the surface is 1 Earth radius ... about 3960 miles.

Gravitational force is inversely proportional to the square of
the distance between the centers of the objects, so in order
to reduce the acceleration of gravity by 1/2, you increase the
distance by √2 .

           (3960 miles) x (√2) = 5,600 miles from the center

                                         = 1,640 miles above the surface.

                                          
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