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3 years ago

During the first stage of sleep deprivation, the subject may __________.

Physics

2 answers:

klio [65]3 years ago

7 0

<h2>First stage of sleep Deprivation Subject </h2>

During the first stage of sleep deprivation, the subject is NREM which stands for non-rapid eye movement. In this condition, we are not sleeping in the depth. It can be said as dreamless sleep. On electro-encephalography recording, the brain waves are not fast so they have high voltage.

In this condition, the breathing, heart rate and blood pressure is low. The sleep is comparatively tranquil. NREM lasts for 90 minutes to 120 minutes. It accounts for about 75% of the normal sleep time. Rapid eye movements do not occur.

Tasya [4]3 years ago

5 0

Answer:

During the first stage of sleep deprivation the subject will become moody, grumpy or irritable.

Explanation:

There are two kinds of sleep deprivation which are classified as acute and chronic sleep deprivation. The acute one is the kind in which patient cannot sleep for one or two nights but in chronic sleep deprivation patient miss the seven hours of sleep per night in a week.

Dr. Ruiz further classified the stages of sleep deprivation as:

Moodiness and irritability.
Difficulty in concentrating.
Difficulty in naming things.
Lapse in memory.
Hallucination.
Paranoia

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A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

vodka [1.7K]

Answer:

a = 603.59 m/s^2

Explanation:

from the data given . the rate of change in magnetic field is as follow

$\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s$

from the faraday's law of induction , the expression for the induced emf in region of radius r as follow

$\epsilon = \frac{d \phi}{dt}$

$\int E.dl = \frac{d(BA)}{dt}$

$E(2\pi r)= \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} \frac{dB}{dt}$

electric field at point P_1 as follow

$E = \frac{r}{2} \frac{dB}{dt}$

$E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}$

$E = 6.3\times 10^{-6} V/m$

from newton 2nd law of motion, the acceleration of proton is

F = ma

qE = ma

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}$

a = 603.59 m/s^2

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3 years ago

Select the correct answer.

elixir [45]

A) Energy is absorbed in the reaction.

Such reactions are called endothermic reactions.

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3 years ago

Read 2 more answers

Need help on these 2 questions please????? Hurry

fredd [130]

Answer:

Question 1 the anwser is 300 meters. Question 2 the answer is speed.

is Explanation:

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4 years ago

A person doing chin-up weighs 700.0 N, disregarding the weight of the arms. During the first 25.0 cm of the lift, each arm exert

dlinn [17]

First, we will get the resultant force:
The direction of the force due to the person's weight is vertically down.
weight of person = 700 newton

Assume that the force exerted by the arms has a vertically upwards direction.
Force exerted by arms = 2*355 = 710 newtons

Therefore, the resultant force = 710 - 700 = 10 newtons (in the vertically upwards direction)

Now, we will get the mass of the person.
weight = 700 newtons
weight = mass * acceleration due to gravity
700 = 9.8*mass
mass = 71.428 kg

Then we will calculate the acceleration of the resultant force:
Force = mass*acceleration
10 = 71.428*acceleration
acceleration = 0.14 m/sec^2

Finally, we will use the equation of motion to get the final speed of the person.
V^2 = U^2 + 2aS where:
V is the final velocity that we need to calculate
U is the initial velocity = 0 m/sec (person starts at rest)
a is the person's acceleration = 0.14 m/sec^2
S is the distance covered = 25 cm = 0.25 meters

Substitute with the givens in the above equation to get the final speed as follows:
V^2 = U^2 + 2aS
V^2 = (0)^2 + 2(0.14)(0.25)
V^2 = 0.07
V = 0.2645 m/sec

Based on the above calculations:
The person's speed at the given point is 0.2645 m/sec

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3 years ago

Read 2 more answers

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before

Darina [25.2K]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}$

$\\$

☯ As we know that,

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

$\\$

$\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }$

$\\$

$\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }$

$\\$

$\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }$

$\\$

$\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\$

$\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }$

$\\$

$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}$

$\\$

$\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}$

$\\$

$\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }$

$\\$

$\dashrightarrow\:\: \sf{ s = 0.04 \: m }$

$\\$

☯ For left penetration (s₂)

$\\$

$\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }$

$\\$

$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

$\\$

$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

$\\$

$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

$\\$

$\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\$

4 0

3 years ago