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worty [1.4K]
3 years ago
12

Some people wish that we lived in a recollapsing universe that would eventually stop expanding and start contracting. For this t

o be the case, which of the following would have to be true (based on current understanding)?)? (A) Dark energy is the dominant form of energy in the cosmos. (B) Dark energy does not exist and there is much more dark matter than we are aware of to date. (C) Neither dark energy nor dark matter really exist. (D) Dark energy exists but dark matter does not.
Physics
1 answer:
Monica [59]3 years ago
4 0

Answer:

(B) Dark energy does not exist and there is much more matter than current evidence suggests.

Explanation:

The repulsive force which is accelerating expansion of the universe is called as dark energy. Most of matter present in the universe is the dark matter of about eighty five percent.

So, a collapsing universe would not have the dark energy and there is more matter which is not the dark matter. This theory is rejected because expansion of the universe is observable.

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!! Offering 50 points !!
Evgesh-ka [11]

Answeryes they are the same

Explanation:

0.5-0.5=10 is gotta be sorry if its wrong

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3 years ago
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Which of the following simple machines is not properly identified? A. Scissors: lever B. Pizza cutter: wedge C. Screw: wheel and
GrogVix [38]

Pretty sure the answer is   C. Screw: wheel and axle

8 0
3 years ago
How does the law of conservation of mass apply to this reaction: C2H4 + O2 → H2O + CO2?
kicyunya [14]

The answer is a

the equation needs to be balanced. There are fewer oxygen atoms in the equation than hydrogen or a carbon

4 0
3 years ago
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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

qV=\frac{1}{2}mv^2

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

K=\frac{1}{2}mv^2

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
What is the current in a 160V circuit if the resistance is 2Ω? V= I= R =
sveta [45]

Answer:

80 amperes

Explanation:

Current in the circuit = ?

Voltage in the circuit = 160 Volts

Resistance = 2 Ω

Voltage = Current x Resistance

V = IR

160V = I x 2 Ω

I = 160V / 2 Ω

I = 80 Amperes

Therefore the current in the circuit is 80 amperes :)

8 0
3 years ago
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