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lianna [129]
3 years ago
5

Yolanda has decided she is going to be a nurse. What is the next step she should take to make her career goals a reality? a. Cre

ate a personal career profile b. Identify her personal values c. Take an interest inventory d. Create an individual career plan
Physics
3 answers:
andrew11 [14]3 years ago
8 0

The correct answer is option D

When someone decides to do something then he or she must prepare themselves according to the career they choose.

When Yolanda has decided to be a nurse then she must create an individual career plan and start working on it.

A goal oriented person only works in the area of their goal and accomplishes it.


Naya [18.7K]3 years ago
6 0

Answer: Option (d) is the correct answer.

Explanation:

Since Yolanda has decided to go for a career in nursing then it is advisable that she should plan an individual career in the same.

This is because a person in whatever field his/her interest is, the person should plan how and what career options he will get after the completion of that specific course in order to earn for his living.

Thus, we can conclude that next step Yolanda should take to make her career goals a reality is that she should create an individual career plan.

randyin2 years ago
0 0

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A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
2 years ago
In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression x= 5.0 cos
lakkis [162]

Answer:

a)   x = 4.33 m ,   b)  w = 2 rad / s ,  f = 0.318 Hz ,  c) a = - 17.31 cm / s²,  

d) T =  3.15 s,  e)  A = 5.0 cm

Explanation:

In this exercise on simple harmonic motion we are given the expression for motion

          x = 5 cos (2t + π / 6)

they ask us for t = 0

a) the position of the particle

      x = 5 cos (π / 6)

      x = 4.33 m

remember angles are in radians

 

b) The general form of the equation is

          x = A cos (w t + Ф)

when comparing the two equations

         w = 2 rad / s

angular velocity and frequency are related

          w = 2π f

           f = w / 2π

           f = 2 / 2pi

           f = 0.318 Hz

c) the acceleration is defined by

      a == d²x / dt²

      a = - A w² cos (wt + Ф)

for t = 0 ,  we substitute

      a = - 5,0 2² cos (π / 6)

      a = - 17.31 cm / s²

d) El period is

          T = 1/f

         T= 1/0.318

         T =  3.15 s

e) the amplitude

        A = 5.0 cm

3 0
3 years ago
What will the pressure of gas be if the temperature rises to 87°C
weeeeeb [17]

Answer: 0 degrees Celsius or 32 degrees Fahrehit

Explanation:

3 0
2 years ago
In the northern hemisphere planetary winds deflect to the
umka2103 [35]

Answer:

Planetary are deflected to right due to Coriolis effect.

Explanation:

   The term Coriolis effect is defined as an effect in which the rotating object experience a force called as coriolis force which acts perpendicular to the axis of rotation and direction of motion. The effect talks about how the moving objects like ocean currents, wind are deflected due to the rotation of earth.

      Winds and ocean currents are strongly affect by this effect.

   

8 0
3 years ago
X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{
dedylja [7]
<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}.c}, being h=4.136(10)^{-15}eV.s the Planck constant, m_{e} the mass of the electron and c=3(10)^{8}m/s the speed of light in vacuum.

\theta=30\° the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at 30\°:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))   (2)

\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m   (3)

Now, the initial energy E_{o}=400keV=400(10)^{3}eV of the photon is given by:

 E_{o}=\frac{h.c}{\lambda_{o}}    (4)

From this equation (4) we can find the value of \lambda_{o}:

\lambda_{o}=\frac{h.c}{E_{o}}    (5)

\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}    

\lambda_{o}=3.102(10)^{-12}m    (6)

Knowing the value of \Delta \lambda and \lambda_{o}, let's find \lambda':

\Delta \lambda=\lambda' - \lambda_{o}

Then:

\lambda'=\Delta \lambda+\lambda_{o}  (7)

\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m  

\lambda'=3.427(10)^{-12}m  (8)

Knowing the wavelength of the scattered photon \lambda'  , we can find its energy E' :

E'=\frac{h.c}{\lambda'}    (9)

E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}    

E'=362.063keV    (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron E_{e}, we have to calculate all the energy lost by the photon, which is:

E_{e}=E_{o}-E'  (11)

E_{e}=400keV-362.063keV  

Finally we obtain the energy of the recoiling electron:

E_{e}=37.937keV  

5 0
3 years ago
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