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Sophie [7]
2 years ago
14

Correlation is:

Physics
1 answer:
xz_007 [3.2K]2 years ago
6 0

Answer: The answer is B

Explanation: A correlation generally is a mutual relationship/connection or the process of establishing the relationship/connection between 2+ things

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How do I convert 14.8 cm into MEters?
stellarik [79]
Multiply it by a fraction equal to ' 1 ', like this:

(14.8 cm) x (1 meter/100 cm) = 14.8/100 = 0.148 meter
6 0
3 years ago
Read 2 more answers
Which term describes the time it takes for an object to complete one full cycle of motion on a spring?
Sav [38]

The term period (symbol: T) describes the time it takes for an object to complete one full cycle of motion on a spring.

The formula for time is: T  = 1 / f , where f is the frequency  , f= c / λ = wave speed c (m/s) / wavelength λ (m)..

The formula describes that as the frequency of a wave increases, the time period of the wave decreases.

4 0
4 years ago
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What is the velocity for an object at rest?
dem82 [27]

Answer:

e

Explanation:

An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

7 0
3 years ago
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The distance from a crest to a trough is called the wave length
V125BC [204]

Answer:

Explanation:

The highest surface part of a wave is called the crest, and the lowest part is the trough. The vertical distance between the crest and the trough is the wave height. The horizontal distance between two adjacent crests or troughs is known as the wavelength.

I hope it's helpful!

5 0
3 years ago
Phosphorus 32 has a half life of 14.3 days how much of a 4mg sample of phosphorus - 32 will remain after 71.5 days
andrew-mc [135]

Let us first find out the radioactive constant of Phosphorus 32.

Radioactive constant, λ = \frac{ln2}{t_{1/2}}

Here, {t_{1/2} is half life of phosphorus 32 = 14.3 days

λ = \frac{ln2}{14.3}

The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,

m=m₀e⁻(λt)

= 4*e^{-\frac{ln2}{14.3}71.5}

= 4*e^{-ln2*5}

=4*e^{-ln(2)^{5}}

=4*e^{-ln32}

=\frac{4}{32}

= 0.125 mg

The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.

6 0
4 years ago
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