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agasfer [191]
3 years ago
6

What is dark energy? How is dark energy related to the theory of the multiverse?

Physics
1 answer:
olga55 [171]3 years ago
4 0

Answer:

In the 1980s, astronomers turned to multiverse theory to explain the “luckily small” amount of dark energy in our universe. According to cosmologist's theories, this small amount of dark energy appeared to enable our universe to host life, while most universes in the multiverse couldn't.

Explanation:

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Most ocean waves obtain their energy and motion from _____.
OLga [1]
Centrifugal force of the earth's rotation as well as tidal forces caused by the moon's orbit and the sun's gravitational pull.
4 0
3 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
Please look at the diagram and ans that question
TEA [102]
I can see that they are running away like my dad did
3 0
2 years ago
When the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium position with a frequenc
Nesterboy [21]

Answer:

w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

Explanation:

We can simulate this system as a physical pendulum, which is a pendulum with a distributed mass, in this case the angular velocity is

          w² = mg d / I

In this case, the distance d to the pivot point of half the length (L) of the cylinder, which we consider long and narrow

         d = L / 2

The moment of inertia of a cylinder with respect to an axis at the end we can use the parallel axes theorem, it is approximately equal to that of a long bar plus the moment of inertia of the center of mass of the cylinder, this is tabulated

        I = ¼ m r2 + ⅓ m L2

        I = m (¼ r2 + ⅓ L2)

now let's use the concept of density to calculate the mass of the system

        ρ = m / V

        m = ρ V

the volume of a cylinder is

         V = π r² L

          m =  ρ π r² L

let's substitute

        w² = m g (L / 2) / m (¼ r² + ⅓ L²)

        w² = g L / (½ r² + 2/3 L²)

        L >> r

         w = √[g /L (½ r²/L2 + 2/3 ) ]

When the mass of the cylinder changes if its external dimensions do not change the angular velocity DOES NOT CHANGE

4 0
3 years ago
A person measures his or her heart rate by counting the number of beats in 30s. If 40±1 beats are counted in 30.0±0.5s, what is
nordsb [41]

Answer:

Rate = 1.33 \pm 0.055 beats per second

Explanation:

Number of heart beats = 40 \pm 1

time taken = 30.0 \pm 0.5 s

now we have

N = 40 \pm 2.5%

t = 30.0 \pm 1.67%

now rate of heart beat is defined as number of heart beat per unit of time

so we have

Rate = \frac{N}{t}

Rate = \frac{40 \pm 2.5}{30 \pm 1.67}

so we have

Rate = 1.33 \pm (2.5 + 1.67 )

Rate = 1.33 \pm 4.17%

Rate = 1.33 \pm 0.055 beats per second

7 0
3 years ago
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