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iris [78.8K]
3 years ago
6

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

of dark energy:
A) recollapsing universe
B) coasting universe
C) critical universe
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).
Physics
1 answer:
Keith_Richards [23]3 years ago
3 0

Answer:

1. Recollapsing universe

2. Critical universe

3. Coasting universe

Explanation:

Recollapsing universe has dark matter density greater than critical density. While critical universe has its matter density equal to the critical sensity. Coasting universe on the other hand has much smaller matter density compared to critical density.

Note that the critical density is approximately 10^-20 grams/cm3

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Mr. Llama walked from his house to the bus stop. The bus stop is 2 miles from his house. He returned back to his house from the
Hoochie [10]

Answer:

Displacement of Mr. Llama: Option D. 0 miles.

Explanation:

The magnitude of the displacement of an object is equal to the distance between its final position and its initial position. In other words, as long as the initial and final positions of the object stay unchanged, the path that this object took will not affect its displacement.

For Mr. Llama:

  • Final position: Mr. Llama's house;
  • Initial position: Mr. Llama's house.

The distance between the final and initial position of Mr. Llama is equal to zero. As a result, the magnitude of Mr. Llama's displacement in the entire process will also be equal to zero.

7 0
3 years ago
When do (object distance) is very large, what does the thin lens equation predict for the value of 1/f?
N76 [4]

Answer:

1 / f = 1 / o + 1 / i

1 / image distance + 1 / object distance = 1  focal length

If the object distance is large the image will be at about the focal length

The value of 1 / f is fixed for any one particular lense

As the object distance decreases the image must increase

The above equation can also be written as

o i / (i + o) = f    or i / (i / o + 1)  = f

If for instance o was very large the image would be at the focal length

6 0
3 years ago
An object has 90000 J of kinetic energy and is moving at 12 m/s. what is the objects mass?
Klio2033 [76]

Answer:

1250kg

Explanation:

we can solve this using the formula for the Kinetic Energy

Kinetic.Energy=\frac{1}{2}*mass*velocity^{2}\\mass=\frac{2*Kinetic.Energy}{velocity^{2}}\\mass=\frac{2*90000}{12^{2}}\\\\mass=1250kg

7 0
4 years ago
Vectors and have scalar product -9.00 and their vector product has magnitude 7.00.
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<span>-(-(-)(-)(-10x))=-5 solve for x</span>
6 0
3 years ago
You push downward on a trunk at an angle 25° below the horizontal with a force of If the trunk is on a flat surface and the coef
Sophie [7]

Complete question is;

You push downward on a trunk at an angle 25° below the horizontal with a force of 750N. if the trunk is on a flat surface and the coefficient of static friction between the surface and the trunk is 0.61, what is the most massive trunk you will be able to move?

Answer:

The most massive trunk is about 81.3 kg

Explanation:

I've attached a free body diagram that depicts this question.

Where;

N = normal force on the trunk

m = mass of the trunk

W = weight of the trunk = mg

F = static frictional force

Using equilibrium of force in vertical direction, we obtain;

N = W + 750 Sin25

N = mg + 750 Sin25    - - - - (eq 1)

Now, we are given that Coefficient of static friction: μ = 0.61

static frictional force is given by the formula;

F = μN

Since N = mg + 750 Sin25, we now have;

F = (0.61) (mg + 750 Sin25)   - - - (eq 2)

Along the horizontal direction, for the trunk to move, force equation must be;

F = 750 Cos25

Thus, we now have;

750 Cos25 = 0.61(mg + 750 Sin25)

g = 9.81.

So,we now have ;

750 Cos25 = 0.61(m(9.81) + 750Sin25)

750 × 0.9063 = 0.61(9.81m + (750 × 0.4226))

Divide both sides by 0.61;

(750 × 0.9063)/0.61 = 9.81m + 316.95

1114.3 = 9.81m + 316.95

1114.3 - 316.95 = 9.81m

797.35 = 9.81m

m = 797.35/9.81

m = 81.3 kg

7 0
3 years ago
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