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4 years ago

15

REM rebound involves the A) tendency for REM sleep periods to become increasingly longer and more frequent as a normal night of

sleep progresses. B) increase in REM sleep that characteristically follows intense learning episodes or stressful daytime experiences. C) unusual symptoms of tiredness and irritability that follow periods of REM sleep deprivation. D) tendency for REM sleep to increase following REM sleep deprivation.

1 answer:

Ivanshal [37]4 years ago

3 0

Answer: i think the answer is D) tendency for REM sleep to increase following REM sleep deprivation.

Explanation:REM rebound is the lengthening and increasing frequency and depth of rapid eye movement (REM) sleep which occurs after periods of sleep deprivation.

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If a machine guard is missing or needs repair, you should ______________.

Gelneren [198K]

B)
Step by step explanation

7 0

3 years ago

Read 2 more answers

Which of the following Python lines calculates the standard deviation of variables Exam1 and Exam4 from a CSV file called ExamSc

Virty [35]

Answer:

The option B calculates the standard deviation of variables Exam1 and Exam4.

Explanation:

First, let's write the first two lines correctly:

import pandas as pd
scores = pd.read_csv('ExamScores.csv')

Now, let's analyse each option:

- The option A is incorrect

The expression std[['Exam1','Exam4']] is the incorrect part because in this case std is working as a variable and the std was not previously declared. So it will generate an error.

- The option C is incorrect

The expression std[['Exam1','Exam4']].scores() is incorrect because of two reasons: first, std is working as a variable and the std was not previously declared (as in option A), second, .scores() is working as a method and in this case, scores is the variable. So it will generate an error.

- The option D is incorrect

The expression [['Exam1']] is the incorrect part because it doesn't include the 'Exam4' and the problem is asking for the standard deviation of variables Exam1 and Exam4.

- The option B is CORRECT

The expression std[['Exam1','Exam4']] refers to the variables Exam1 and Exam4 from the total exams. As scores is a pandas variable and .std() is a method provided by pandas library, the expression .std() allows to get the deviation standard of the variables Exam1 and Exam4.

Thus, the option B calculates the standard deviation of variables Exam1 and Exam4.

4 0

3 years ago

Air at 1600 K, 30 bar enters a turbine operating at steady state and expands adiabatically to the exit, where the pressure is 2.

djyliett [7]

Solution :

The isentropic efficiency of the turbine is given as :

$\eta = \frac{\text{actual work done}}{\text{isentropic work done}}$

$=\frac{m(h_1-h_2)}{m(h_1-h_{2s})}$

$=\frac{h_1-h_2}{h_1-h_{2s}}$

The entropy relation for the isentropic process is given by :

$0=s^\circ_2-s^\circ_1-R \ln \left(\frac{P_2}{P_1}\right)$

$\ln \left(\frac{P_2}{P_1}\right)=\frac{s^\circ_2-s^\circ_1}{R}$

$\frac{P_2}{P_1}=exp\left(\frac{s^\circ_2-s^\circ_1}{R}\right)$

$\left(\frac{P_2}{P_1}\right)_{s=constant}=\frac{P_{r2}}{P_{r1}}$

Now obtaining the properties from the ideal gas properties of air table :

At $T_1 = 1600 \ K,$

$P_{r1}=791.2$

$h_1=1757.57 \ kJ/kg$

Calculating the relative pressure at state 2s :

$\frac{P_{r2}}{P_{r1}}=\frac{P_2}{P_1}$

$\frac{P_{r2}}{791.2}=\frac{2.4}{30}$

$P_{r2}=63.296$

Obtaining the properties from Ideal gas properties of air table :

At $P_{r2}=63.296$ , $T_{2s}\approx 860 \ K$

Considering the isentropic relation to calculate the actual temperature at the turbine exit, we get:

$\eta=\frac{h_1-h_2}{h_1-h_{2s}}$

$\eta=\frac{c_p(T_1-T_2)}{c_p(T_1-T_{2s})}$

$\eta=\frac{T_1-T_2}{T_1-T_{2s}}$

$0.9=\frac{1600-T_2}{1600-860}$

$T_2= 938 \ K$

So, at $T_2= 938 \ K$ , $h_2=975.66 \ kJ/kg$

Now calculating the work developed per kg of air is :

$w=h_1-h_2$

= 1757.57 - 975.66

= 781 kJ/kg

Therefore, the temperature at the exit is 938 K and work developed is 781 kJ/kg.

4 0

3 years ago

A cylindrical specimen of a titanium alloy having an elastic modulus of 107 GPa (15.5 × 106 psi) and an original diameter of 3.7

Keith_Richards [23]

Answer:

the maximum length of specimen before deformation is found to be 235.6 mm

Explanation:

First, we need to find the stress on the cylinder.

Stress = σ = P/A

where,

P = Load = 2000 N

A = Cross-sectional area = πd²/4 = π(0.0037 m)²/4

A = 1.0752 x 10^-5 m²

σ = 2000 N/1.0752 x 10^-5 m²

σ = 186 MPa

Now, we find the strain (∈):

Elastic Modulus = Stress / Strain

E = σ / ∈

∈ = σ / E

∈ = 186 x 10^6 Pa/107 x 10^9 Pa

∈ = 1.74 x 10^-3 mm/mm

Now, we find the original length.

∈ = Elongation/Original Length

Original Length = Elongation/∈

Original Length = 0.41 mm/1.74 x 10^-3

<u>Original Length = 235.6 mm</u>

5 0

3 years ago

Going too slow on the band will ______

vlabodo [156]

Answer:

,reduce accuracy, relief, 1/4, • push stock with hands towards the blade of band saw

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZ

4 0

3 years ago