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inysia [295]
3 years ago
7

How does the change in the temperature of the universe provide evidence for universe expansion that supports the Big Bang Theory

? (1 point)
The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.


The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.​
Physics
1 answer:
xeze [42]3 years ago
8 0

Question:

1) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.

2) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

3) The universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses.

4) The universe is warming which, according to the Big Bang Theory, is expected to happen as the cosmos accumulates.​

Answer:

The correct option is;

3) The Universe is cooling which, according to the Big Bang Theory, is expected to happen as the cosmos disperses

Explanation:

With the temperature measurement carried out using the CSIRO radio telescope, Astronomers have been able to determine a temperature difference in the universe from 5.08 Kelvin 7.2 billion light years away to 2.73 Kelvin in the Universe today, which is in support of the Big Bang theory that as the Universe expanded from a state of extreme temperature that cools down as the Universe expands or the cosmos disperses.

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A 1000 kg race car rounds a curve with a radius of 50 m.
Kruka [31]

Answer:

5644556677888777766554433

5 0
2 years ago
Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w
gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

u(x) = \frac{-2}{x} + \frac{4}{x^2}

At x = 1.00 m

u(1) = \frac{-2}{1} + \frac{4}{1^2}

= 4J

Kinetic energy = (1/2)mv²

= \frac{1}{2} * 4(3)^2

= 18J

Total energy will be =

4J + 18J = 22J

At x = 5

u(5) = \frac{-2}{5} + \frac{4}{5^2}

= \frac{4-10}{25} = \frac{-6}{25} J

= -0.24J

Kinetic energy =

\frac{1}{2} * 4Vf^2

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

Initial total energy = final total energy

22 = 2Vf² - 0.24

Vf² = (22+0.24) / 2

Vf = \sqrt{frac{22.4}{2}

= 3.33 m/s

b) magnitude of force when x = 5.0m

u(x) = \frac{-2}{x} + \frac{4}{x^2}

\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}

= \frac{2}{x^2} - \frac{8}{x^3}

At x = 5.0 m

\frac{2}{5^2} - \frac{8}{5^3}

F = \frac{2}{25} - \frac{8}{125}

= 0.016N

8 0
3 years ago
A circuit can only light up a lightbulb if there is a ______ path for electricity to travel from one end of the energy source to
Dmitry_Shevchenko [17]

Answer:

Continuous

Explanation:

A circuit can only light up a lightbulb if there is a continuous path for electricity to travel from one end of the energy source to the other end.

4 0
3 years ago
Read 2 more answers
a wooden block has a mass of 1.2 kg, a specific heat of 710, and is at a temperature of 25*C. what is the block's final temperat
mash [69]

The final temperature of the block is 27.5^{\circ} C

Explanation:

The amount of thermal energy Q supplied to a substance is related to the increase in temperature of the substance, \Delta T, according to the equation

Q=mC_s \Delta T

where:

m is the mass of the substance

C_s is the specific heat capacity of the substance

In this problem, we have:

m = 1.2 kg is the mass of the block

Q = 2,130 J is the amount of energy supplied to the block

C_s = 710 J/kg^{\circ}C is the specific heat capacity of the block

Solving for \Delta T, we find the increase in temperature:

\Delta T = \frac{Q}{m C_s}=\frac{2130}{(1.2)(710)}=2.5^{\circ}C

And since the initial temperature was

T_i = 25^{\circ}C

The final temperature will be

T_f = T_i + \Delta T = 25+2.5=27.5^{\circ} C

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

8 0
3 years ago
Galileo attempted to measure the speed of light by measuring the time elapsed between his opening a lantern and his seeing the l
Dafna11 [192]

Answer:

The distance is 2.58\times10^{7}\ m.

Explanation:

Given that,

Time \Delta t = 0.2\ s

The velocity is no more than a 14 % error in the speed of light.

So,

Velocity v= c\times 86\%

We need to calculate the distance

Using formula of speed

v = \dfrac{d}{t}

d = v\times t

Where, v = speed

d = distance

t = time

Put the value into the formula

d = 3\times10^{8}\times\dfrac{86}{100}\times0.2

d=516\times10^{5}\ m

We know that,

The one side distance d' is

d'=\dfrac{d}{2}

d'=\dfrac{516\times10^{5}}{2}

d'=2.58\times10^{7}\ m

Hence, The distance is 2.58\times10^{7}\ m.

5 0
3 years ago
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