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3 years ago

What is the main promblem with survey research?

Physics

1 answer:

zlopas [31]3 years ago

7 0

Answer:

A major problem in all survey research is that respondents are almost always self-selected. Not everyone who receives a survey is likely to answer it, no matter how many times they are reminded or what incentives are offered.

Explanation:

<em><u>DISADVANTAGES</u></em>

Respondents may not feel encouraged to provide accurate, honest answers.

Respondents may not feel comfortable providing answers that present themselves in a unfavorable manner.

Respondents may not be fully aware of their reasons for any given answer because of lack of memory on the subject, or even boredom.

HAVE A GOOD DAY!

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The acceleration vector of a particle in projectile motion ________.

Alex73 [517]

Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

4 0

3 years ago

A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 4500 N/C and is directed in

Anton [14]

Answer with Explanation:

We are given that

$E=4500 N/C$

$q=-5.5\times 10^{-9} C$

a.x=-0.2 m

$E'=\frac{Kq}{r^2}=\frac{9\times 10^9\times 5.5\times 10^{-9}}{(0.2)^2}=1237.5 N$

Where $k=9\times 10^9$

Net electric field=E+E'=4500+1237.5=5737.5 N/C

b.x=0.20 m

Net electric field=E-E'=4500-1237.5=3262.5 N/C

c.y=0.20 m

E=1237.5 N/C

Net electric field= $\sqrt{E^2+E'^2}=\sqrt{(4500)^2+(1237.5)^2}=4667.05 N/C$

7 0

3 years ago

Displacement vectors of 3 m and 5 m in the same direction combine to make a

n200080 [17]

Displacement vector that is 8 im magnitude

5 0

4 years ago

Three identical balls are thrown from the same height off a tall building. They are all thrown with the same speed, but in diffe

koban [17]

Answer:

three balls have the same speed

Explanation:

In a parabolic motion we have that

$v_{x}=v_{0}cos\alpha\\v_{y}=-gt+v_{0}sin\alpha\\v=\sqrt{v_{x}^{2}+v_{y}^{2}}\\$

but the time just before the balls hit the ground is

$t=\frac{2v_{0}sin\alpha}{g}$

Hence we have

ball A

$v=\sqrt{v_{0}^{2}cos^{2}(0)+(-g\frac{2v_{0}sin(0)}{g}+v_{0}sin(0))^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball B

$v=\sqrt{v_{0}^{2}cos^{2}(45)+(-g\frac{2v_{0}sin(45)}{g}+v_{0}sin(45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

ball C

$v=\sqrt{v_{0}^{2}cos^{2}(-45)+(-g\frac{2v_{0}sin(-45)}{g}+v_{0}sin(-45))^{2}}\\v=\sqrt{v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}+v_{0}^{2}(\frac{\sqrt{2}}{2})^{2}}\\v=\sqrt{v_{0}^{2}}=v_{0}$

Hence, all three balls have the same speed just before hit the groug

vA=vB=vC

I hope this is useful for you

regards

5 0

3 years ago

Computers have become a familiar ___ of communication, connecting people around the world.

devlian [24]

They're a medium (c) of communication

A. mediation is what you so when you want to bring the sides of conflict together

b. is a person who is advocating something

d. is a presentation of some opinion or facts.

5 0

3 years ago