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3 years ago

Hans Selye’s general adaptation syndrome theory proposes that adaptation to stress occurs in how many stages?

Physics

1 answer:

ANTONII [103]3 years ago

4 0

<u>Answer:</u>

Adaption to stress occurs in three stages: alarm, fight or flight, exhaustion.

<u>Explanation:</u>

According to the general adaptation syndrome theory proposed by Hans Selye, the adaption to stress occurs in three stages which are:

1. alarm

2. fight or flight

3. exhaustion

This is a process which comprises of three stages that describes the physiological changes which a body undergoes when in stress (an emotional, mental and physical human response to a specific stimulus).

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Determine the magnitude and direction of the resultant force of the following free body diagram.

Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force ( $\vec F$ ), in newtons, by vectorial sum:

$\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}$ (1)

$\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}$

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

$\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}$

$\|\vec F\| \approx 599.923\,N$

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

$\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)$

Where $\theta$ is the direction of the resultant force, in sexagesimal degrees.

$\theta \approx 36.405^{\circ}$

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0

3 years ago

What are the damages or effects of sewage dumps into natural water sources???

masha68 [24]

Answer:

Sewage dumps pollute our waterways causing death of marine life, water pollution, and disease.

Explanation:

Sewage dumps. bring disease or death to our ecosystems

5 0

3 years ago

Read 2 more answers

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

3 years ago

A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance.

Nookie1986 [14]

Answer:

a. 0.6 m b. 0.385 m c. 3.6 m/s at 287.78° to the horizontal

Explanation:

a. Using s = ut - 1/2gt² for motion under gravity where s = vertical distance = height of table, u = initial vertical velocity of book = 0 m/s, t = time of flight = 0.350 s and g = acceleration due to gravity = 9.8 m/s².

Substituting these these values into s and taking the top of the table as position 0 m, we have.

0 - s = 0t - 1/2gt²

-s = -1/2gt²

s = 1/2gt²

s = 1/2 × 9.8 m/s² × (0.350 s)²

s = 0.6 m

b. Using d = v't where d = horizontal distance from table, v' = horizontal velocity of book = 1.10 m/s and t = time of flight = 0.350 s

d = v't = 1.10 m/s × 0.350 s = 0.385 m

c. Using v² = u² - 2gs where u = initial vertical velocity of book = 0 m/s and g = 9.8 m/s², s = -0.6 m (negative since we are at the bottom and 0 m is at the top)and v = final vertical velocity of book

v² = u² - 2gs

= 0 - 2 × 9.8 m/s² × (-0.6 m)

= 11.76 m²/s²

v = √11.76 m/s

= 3.43 m/s

So, the magnitude of the resultant velocity is V = √(v² + v'²)

= √((3.43 m/s)² + (1.10 m/s)'²)

= √(11.76 m²/s² + 1.21 m²/s²)

= √12.97 m²/s²

= 3.6 m/s

Its direction Ф = tan⁻¹(-v/v') since v is in the negative y direction

= tan⁻¹(-3.43 m/s/1.10 m/s)

= tan⁻¹(-3.1182)

= -72.22°

Ф = -72.22°+ 360 = 287.78° since it is in the third quadrant

7 0

3 years ago

Care este pozitia unui mobil cand t(s)=14 si x(m)=12?

mario62 [17]

name shrawan sha

DtD_LpOZ3Xm_0ff

3 0

3 years ago