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bixtya [17]
3 years ago
12

How do we determine the conditions that existed in the very early universe? A We can only guess at the conditions, since we have

no way to calculate or observe what they were. B The conditions in the very early universe must have been much like those found in stars today, so we learn about them by studying stars. C From the current expansion rate we can work backward to estimate temperature and densities at various times in the early universe. D By looking all the way to the cosmological horizon, we can see the actual conditions that prevailed all the way back to the first instant of the Big Bang.
Physics
1 answer:
lakkis [162]3 years ago
5 0

Answer:

D By looking all the way to the cosmological horizon, we can see the actual conditions that prevailed all the way back to the first instant of the Big Bang.

Explanation:

Astrophysicists are able to determine the conditions that existed in the early universe, by using instruments such as telescopes to observe and study cosmic horizons. More ideas about the early universe can be found from the thermal light present in cosmic backgrounds.

Scientists study these details that provide an insight into the conditions that existed so many years ago. They have been able to determine that the Big Bang involved so many collisions from these observations.

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Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T
ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

                   \vec{A}=x_A\hat{i}

Vector B , which has a magnitude of 5.5 m

                   \vec{B}=x_B\hat{i}+y_B\hat{j}

                   \sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A                    \vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}

Comparing we will get

                  x_A=-x_B\\\\y_B=6x_A

Substituting in x_{B}^{2}+y_{B}^{2}=30.25

                  \left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904

So we have

    \vec{A}=0.904\hat{i}

Magnitude of vector A = 0.904

8 0
3 years ago
The particle is an electron. the field slows down the electron without deflecting it. what is the direction of the electric fiel
sammy [17]

The particle is an electron. The field slows down the electron without deflecting it. The direction of the electric field is <u>right.</u>

In physics, the motion of electrically charged particles gives rise to a field called an electric field. It is measured in force per unit charge.

This field applies force on other charged particles.

Particles bearing opposite charges attract each other while particles having similar charges repel each other in the field.

If a positive charge is placed in the field then the field line moves in an outward direction and for a negative charge, the direction of the lines is inward.

If you need to learn more about electric field click here:

brainly.com/question/26199225

#SPJ4

7 0
2 years ago
If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?
sweet [91]

Answer:

slbohohohohhoohhoohooohhoohho

6 0
2 years ago
What type of motion occurs when an object spins around an axis without altering its linear position?
dimulka [17.4K]
That would be only rotational motion
5 0
3 years ago
A current of 2.0A flows through a light bulb. What is the amount of charge flowing through the light bulb in 40 seconds?
lidiya [134]

Answer:

Quantity of charge = 80 Coulombs

Explanation:

Given the following data;

Current = 2 A

Time = 40 seconds

To find the amount of charge flowing through the light bulb;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge = current * time

Substituting into the formula, we have;

Quantity of charge = 2 * 40

Quantity of charge = 80 Coulombs

5 0
3 years ago
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