All categories

3 years ago

A correlation coefficient represents what two things

Physics

1 answer:

kenny6666 [7]3 years ago

3 0

Answer:

A correlation coefficient represents the following:

1- The direction of the relationship

2- The strength of the relationship

You might be interested in

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Juli2301 [7.4K]

Answer:

0.0984

Explanation:

From the first diagram attached below; a free flow diagram shows the interpretation of this question which will be used to solve this question.

From the diagram, the horizontal component of the force is:

$F_X = F_{cos \ \theta}$

Replacing 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

On the other hand, the vertical component is ;

$F_Y = Fsin \ \theta$

Replacing 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

However, resolving the vector, let A be the be the component of the mutually perpendicular directions.

The magnitude of the two components is shown in the second attached diagram below and is now be written as A cos θ and A sin θ

The expression for the frictional force is expressed as follows:

$f = \mu \ N$

Where;

$\mu$ is said to be the coefficient of the friction

N = the normal force

Similarly the normal reaction (N) = mg - F sin θ

Replacing $F_Y \ for \ F_{sin \ \theta}$ . The normal reaction can now be:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals to frictional force.

The horizontal component of the force is given as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Making $\mu$ the subject of the formular in the above equation; we have the following:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Replacing the following values: i.e

$F_X \ = \ 64.65 \ N$

m = 73 Kh

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Then:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Thus, the coefficient of friction is = 0.0984

5 0

3 years ago

in a lab, resonance tubes are used to determine experimentally the speed of sound. using the data given, evaluate the best appro

jek_recluse [69]

The approximate speed of sound is about 761 mph or 335.28 meters/second.

<h3>What's the approximate speed of sound?</h3>

If we consider the normal atmosphere at sea level, the speed of sound is about 761 mph or 335.28 meters/second. The speed of sound is different in different mediums such as solid and liquid.

So we can conclude that the approximate speed of sound is about 761 mph or 335.28 meters/second.

Learn more about speed here: brainly.com/question/4931057

#SPJ1

3 0

2 years ago

Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay abo

Luba_88 [7]

Answer:

#_time = 7.5 10⁴ s

Explanation:

In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.

t = $\frac{t_p}{ \sqrt{1- (v/c)^2} }$

where t_p is the person's own time in an immobile reference frame,

$t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }$

let's calculate

we assume that the speed of the space station is constant

$t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }$

$t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}$

t_ = 0.99998666657 s

therefore the time change is

Δt = t - t_p

Δt = 1 - 0.9998666657

Δt = 1.3333 10⁻⁵ s

this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s

#_time = 1 / Δt

#_time = $\frac{1}{1.3333 \ 10^{-5}}$