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Luba_88 [7]
3 years ago
15

Suppose you're studying levels of job stress and measures of job efficiency in a nursing home. Which one of the following statem

ents best represents a data comparison that demonstrates a positive correlation
A. As stress levels increase, job efficiency increases.
B. As job efficiency decreases, stress levels increase.
C. As job efficiency increases, stress levels decline.
D. As stress levels increase, job efficiency declines
Physics
2 answers:
zloy xaker [14]3 years ago
7 0

Answer: Hello mate!

You are studying the relationship between levels of job stress and measures of job efficiency.

now, we want to describe a positive correlation.

A correlation between two quantities, A and B, is A = kB

where k is a constant.

If the correlation is positive, means that when A increases also increase B, and then k is a positive number.

This means that, in our case,  when job stress increases, job efficiency also increases.

Then the right answer is A "As stress levels increase, job efficiency increases."  

coldgirl [10]3 years ago
6 0
B. As job efficiency increases, stress levels decline (THIS IS WRONG)I think its between A. As job efficiency decreases, stress levels increaseD. As stress levels increase, job efficiency increases
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6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t
zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

d=v\times t

d=9.1\times t....(I)

The distance of longitudinal wave

d=5.7\times (t+71)....(II)

From the first equation

t=\dfrac{d}{9.1}

Put the value of t in equation (II)

d =5.7\times(\dfrac{d}{9.1}+71)

\dfrac{9.1d-5.7d}{9.1}=71\times5.7

d0.3736=404.7

d =1083.24\ km

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0
3 years ago
g The bottom end of a long vertical tube filled with liquid is opened in a basin exposed to air having pressure 100.8 kilo-Pasca
Masja [62]

Answer:

979.6 kg/m³

Explanation:

We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²

So, density ρ = P/hg

Since P = 100.8 kPa = 100.8 × 10³ Pa

substituting the values of the variables into the equation for ρ, we have

ρ = P/hg

= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)

= 100.8 × 10³ Pa ÷ 102.9 m²/s²

= 0.9796 × 10³ kg/m³

= 979.6 kg/m³

So, the density of the liquid is 979.6 kg/m³

3 0
3 years ago
A student connects a small solar panel to a 40 a resistor to make a simple circuit. The solar panel produces a voltage of 2 0.
Nastasia [14]
<h3>Solution for the above question : -</h3>

Ohm's law states that :

  • v = ir

the terms used are :

  • r = resistance
  • v = potential \:  \: difference
  • i  = current

let's solve for electric current :

  • 2 = i \times 40

  • i =  \dfrac{2}{40}

  • i = 0.05 \: A

  • i = 50 \: mA

\mathfrak{good\:  \: luck \:  \: for \:  \: your \:  \: assignment}

8 0
3 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
Read 2 more answers
An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potentia
Vinvika [58]

Answer:

-8.56V

Explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

We can calculate the voltage across the circuit with the emf formula, that is,

e(t) = e* sin(wt)

e(t) = 6.04 * sin(Φ + π)

e(t) = 6.04 * sin(32.5 + 180)

e(t) = -3.245 V

Now, Using Kirchoff Voltage Law,

e(t) - VR- VL - VC = 0

-3.24 - 0 - VL - 5.32 = 0

Finally we have the potential difference across the inductor.

VL = - 8.56 v

5 0
3 years ago
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