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Maksim231197 [3]
3 years ago
6

The process of examining a change in one variable in a model while assuming that all the other variables remain constant is call

ed _______.
Physics
1 answer:
mixer [17]3 years ago
8 0

Answer:

<em>Ceteris Paribus</em>

Explanation:

The process of examining a change in one variable in a model while assuming that all the other variables remain constant is called <em><u>Ceteris Paribus</u></em>.

<em>Ceteris Paribus</em> is a Latin phrase that means "all other things being equal" or "all other things held constant" in English. The phrase has found application in disciplines like Economics and Statistics. This phrase as being adopted as a process of examining a change in one variable in a model while assuming that all the other variables remain constant to ascertain the relationship between the variables or make deductions from an experimental study. An example of <em>Ceteris Paribus</em> application is the law of demand and supply in Economics. The law of demand states, <em>Ceteris Paribus</em>, the higher the price, the lower the quantity demanded and <em>vice versa. </em>Conversely, the law of supply states, <em>Ceteris Paribus</em>, the higher price, the higher the quantity supply and <em>vice versa</em>.

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An Olympic runner completes the 200-meter sprint in 23 seconds. What is the runner’s average speed? (Round your answer to the ne
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A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un
Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

3 0
3 years ago
The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7
My name is Ann [436]

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

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3 years ago
Why do organisms need nitrogen
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