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3 years ago

Explain the theory Steven Reiss developed and tested.

2 answers:

7 0

Steven Reiss proposed the theory of 16 basic desires (under theory of motivation) which he thought of when he was admitted in hospital and visualized the nurses enjoying their work.

He deduced 16 desires which are as follows –

1. Being curious

2. Being hungry

3. Being accepted

4. Taking care of offsprings

5. Being faithful

6. Requiring social justice

7. Being independent

8. Need for establishes environment

9. Need of work out

10. Will of power

11. Romance

12. Desire to save

13. Desire to establish social contact

14. Desire to feel safe and secure

15. Desire to establish oneself in society

16. Desire to revenge

NNADVOKAT [17]3 years ago

6 0

A theory of motivation by Steven Reiss, the 16 Basic Desires Theory talks about the sixteen fundamental needs, values and drives that motivate a person.

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If 7 ounces of a sports drink contains 110 milligrams of sodium, what is the total number of milligrams of sodium in 20 ounces o

ehidna [41]

so you just take 110 divided by 7 and then you get the answer and times tthat by 20 and you get you answer which is 314.28 milligrams of sodium in 20 ounces of the sports drink.

7 0

2 years ago

Steel Usually forms a

meriva

Answer:

Permanent magnetism (of the steel)

make me brainliestt :))

6 0

3 years ago

an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall, what is the height of

Hitman42 [59]

The height of the object will be -5.19 cm

A concave mirror's reflecting surface curves inward and away from the light source. Light is reflected inward to a single focus point via concave mirrors. Concave mirrors, in contrast to convex mirrors, produce a variety of images depending on the object's to the mirror.

Given an object 24.0 cm from a concave mirror creates a virtual image at -33.5 cm. if the image is 7.25 cm tall

So let,

v = Image distance from the mirror = -33.5 cm

u = object distance from the mirror (concave) = 24 cm

hi = Image height = 7.25 cm

h = height of the object = ?

Using below formula to find height of the object

-v/u = hi/h

Putting all value in the formula we get

-(-33.5)/(-24) = 7.25/h

h = -5.19 cm

Therefore the height of the object will be -5.19 cm

Learn more about Concave mirror here:

brainly.com/question/3727024

#SPJ10

3 0

1 year ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

garri49 [273]

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

$2gh = v_f^2 - v_i^2$

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

$(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\$

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

$v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s$

(c)

Now we will consider the downward motion and use the third equation of motion:

$2gh = v_f^2-v_i^2$

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

$2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\$

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

$v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s$

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

7 0

2 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago