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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:

DISADVANTAGES

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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In order to determine the velocity, you must know

Naddik [55]

You need to know the speed and direction of object

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3 years ago

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sara and tory are out fishing on the lake on a hot summer day when they both decide to go for a swim. sara dives off the front o

crimeas [40]

Here it is an application of Newton's III law

as we know by Newton's III law that every action has equal and opposite reaction

So here as we know that two boys jumps off the boat with different forces

from front side of the boat the boy jumps off with force 45 N which means as per Newton's III law if boy has a force of 45 N in forward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 45 N

Now similar way we can say

from back side of the boat the boy jumps off with force 60 N which means as per Newton's III law if boy has a force of 60 N in backward direction then he must apply a reaction force on the boat in reverse direction of same magnitude

So boat must have an opposite force on front end with magnitude 60 N

So here net force due to both jump on the boat is given by

$F_{net} = F_1 - F_2$

$F_{net} = 60 - 45$

$F_{net} = 15 N$

so boat will have net force F = 15 N in forward direction due to both jumps

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3 years ago

A very long, uniformly charged cylinder has radius R and linear charge density λ. Find the cylinder's electric field strength ou

mixer [17]

The cylinder's electric field magnitude, at a distance r from the axis of the cylinder (greater than the cylinder's radius), is equal to $E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

<h3>Further explanation</h3>

Matter is the building block of everything that we encounter in our lives. Matter is made of atoms, which are in turn made of tiny particles which are called electrons, protons, and neutrons. The ammount of these 3 elements, and their topological configuration in the atoms, is what determines what a certain element is (like Carbon, Hydrogen, Iron, etc).

In some cases, some elements may lose or gain some electrons. Regarded that this missing or extra electrons are not very high in number, the material doesn't lose any of its properties, however it will always try to get its number of electrons back to normal. This is when we say that an element has a charge, which is a measure of how much electrons a body needs to get back to normal. A body has positive charge if it lacks electrons, and has negative charge if it has extra electrons.

This charge causes the material to have an Electric field, which is a measure of how much does it attract or repel electrons. In the case of our problem, we need to compute exactly that, the Electric field. In our problem, we have an infinitely long cylinder with a linear charge density $\lambda$ , this means that all parts of the cylinder have the same charge, and due to symmetry, the electric field is constant on the angular and longitudinal directions of the cylinder.

This makes easy to apply Gauss' Law, since for a Gaussian curve in the shape of a concentric cylinder (with a higher radius than that of our charged cylinder) we can write:

$\Phi = \frac{\lambda \cdot L}{\epsilon_0}$

Where $\Phi$ is called the Electric flux. Since the electric field is constant for a given distance r from the axis of the cylinder we can write that:

$\Phi = E \cdot 2\pi r \cdot L$

Joining both our expressions we can get that:

$E= \frac{\lambda}{2\pi \epsilon_0 \cdot r}$

<h3 /><h3>Learn more</h3>

Description on Electric fields: brainly.com/question/8971780
Relation between electric fields and magnetism: brainly.com/question/2838625
How can we use electric charges: brainly.com/question/10427437

<h3>Keywords</h3>

Electrons, protons, electric field, cylinder, electric flux

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e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

anastassius [24]

The force acting between the particles is

$F=k \frac{Q_{1}Q_{2}}{r^2}$
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3 years ago

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The speed of a wave in a medium is effected by____,____, and____

lisov135 [29]

$\huge\mathfrak\green{answer}$

As per as my knowledge

The speed of a wave in a medium is affected by density, wavelength and temperature :)

(Good luck on your test and mark me brainliest if this helps)

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3 years ago