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navik [9.2K]
3 years ago
5

Which of the following is MOST needed for cosmologists to study the age of the universe? A. Energy levels B. Static C. Size D. D

istance traveled
Physics
2 answers:
dedylja [7]3 years ago
6 0
I think the one that is most needed for cosmologist to study the age of the universe is : D. distance travelled
It is used to determine the Hubble constant  today and extrapolating back in time with the observed value of density parameters

Hope this helps


Trava [24]3 years ago
3 0

The last person is wrong, I just took the test and the correct answer was: B.Static

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Another word for character in sports is "sportsmanship." —<br><br> True<br><br> False
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Answer:

False

Explanation:

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It takes 4 hr 39 min for a 2.00-mg sample of radium-230 to decay to 0.25 mg. What is the half-life of radium-230? A) 1 hr 4 min
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In music, the note (A) above middle C has a frequency of 440 Hz. Suppose that note is played in the air where the velocity of so
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5 0
4 years ago
A level of
Wittaler [7]

Answer:

x = 2 meters.

Explanation:

Let the position (distance) of fulcrum to the load be x.

Given the following data;

Load = 40 kg

Effort (force) = 40 Newton

Effort arm = 4 - x

To find the position of the fulcrum, we would use the expression;

Effort * effort arm = load * load arm

40 * (4 - x) = 40 * x

160 - 40x = 40x

160 = 40x + 40x

160 = 80x

x = 160/80

x = 2 meters

Therefore, the position (distance) of fulcrum to the load is 2 meters.

7 0
3 years ago
Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength
Vikentia [17]

Answer:

\frac{E_{A}}{E_{B}}=4

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

E=\frac{F}{q}.

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

F=\frac{kQq}{r^{2}}.

By substitution we get that

E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}

Now, letting E_{A} be the electric field at point A, letting E_{B} be the electric field at point B, and letting R be the distance from the charge to A:

E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}.

The ration of the electric fields is

\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4

This means that at half the distance, the electric field is four times stronger.

4 0
3 years ago
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