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Likurg_2 [28]
3 years ago
9

Describe effective communication strategies for gathering information, educating patients

Physics
1 answer:
timama [110]3 years ago
6 0

Explanation:

Effective communication with patients will enable one to know the needs of the patient better as well as reducing the barriers to understanding each other for both parties.

To be an effective communicator while educating patients, the person must:

  • It is important to establish good rapport with the patient. By so doing they can trust you and let you in. Show empathy. Do not make them feel like you are judging them

  • Use proper body language. Make eye contacts and try to be on the same level as the patient so you can be face to face with them.
  • make the interaction easier for them. You have to keep questions as well as your sentences short and moderate. Stay on topic and always make sure that concepts are clear to them.
  • show respect. try not to speak with commands. Give the patient opportunity to make choices.
  • be patient with them. Due to age or the nature of their illnesses, the patient may be slow in speech or movement. help them to move at their own pace by not rushing them.
  • give them time to respond and ask questions. this will make communication more effective.
  • you cause graphics where necessary or written instructions for the patient.
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An object is dropped at a height of 12 m from the ground. How fast is it moving just before it hits the ground?
sp2606 [1]

The final speed of the object is 15.3 m/s

Explanation:

The object in the problem is moving by uniformly accelerated motion (it is in free fall), so we can use the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a=g=9.8 m/s^2 is the acceleration of gravity (taking downward as positive direction)

s is the vertical displacement

For the object in this problem:

u = 0 (it starts from rest)

s = 12 m is the displacement of the object

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(9.8)(12)}=15.3 m/s

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6 0
3 years ago
a car traveling north with a velocity of 29m/s changes its velocity to 36 in the span of 11s. What was the car's acceleration du
Nikolay [14]

Answer:

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8 0
3 years ago
The pressure, P, of a gas varies directly with its temperature, T, and inversely with its volume, V, according to the equation f
lbvjy [14]
The equation formula:
P V = n R T
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7 0
3 years ago
Read 2 more answers
Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
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