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!
Answer:
a
The radial acceleration is 
b
The horizontal Tension is 
The vertical Tension is 
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is 
The mass of the bob is 
The angle made by the string is 
The centripetal force acting on the bob is mathematically represented as

Now From the diagram we see that this force is equivalent to
where T is the tension on the rope and v is the linear velocity
So

Now the downward normal force acting on the bob is mathematically represented as

So

=> 
=> 
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

=> 
substituting values


The horizontal component is mathematically represented as

substituting value

The vertical component of tension is

substituting value

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

substituting value
![T = [(0.3294) i + (3.3712)j ] \ N](https://tex.z-dn.net/?f=T%20%20%3D%20%5B%280.3294%29%20i%20%20%2B%20%283.3712%29j%20%5D%20%5C%20%20N)
Chemicals are released by the damaged tissue to bring platelets to the area to become sticky and to plug the vessels. An inflammatory response occurs which brings immune cells to fight infection and other clotting factors come to begin forming a new tissue.
Answer:
a) 
Now we can replace the velocity for t=1.75 s

For t = 3.0 s we have:

b) 
And we can find the positions for the two times required like this:
And now we can replace and we got:

Explanation:
The particle position is given by:

Part a
In order to find the velocity we need to take the first derivate for the position function like this:

Now we can replace the velocity for t=1.75 s

For t = 3.0 s we have:

Part b
For this case we can find the average velocity with the following formula:

And we can find the positions for the two times required like this:
And now we can replace and we got:

Answer:
6 N
Explanation:
= Mass flow rate = 1 kg/s
v = Final velocity = 6 m/s
u = Initial velocity = 0 m/s
Force is obtained when we divide change in momentum by time

The force the person exert on the extinguisher in order to prevent it from accelerating is 6 N