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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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cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?

serg [7]

The heat moves from the hot chocolate to the handle of the spoon by a process called thermal conduction.

It is the transfer of heat energy from one object to another when they are in contact with eachother.

Hope this answers your question.

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3 years ago

A fly sits on a potter's wheel 0.30 m from its axle. The wheel's rotational speed decreases from 4.0 rad/s to 2.0 rad/s in 5.0 s

Likurg_2 [28]

Answer: The wheel's average rotational acceleration is -0.4 radians per second squared (rad/s^2)

Explanation: Please see the attachments below

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3 years ago

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Phobos's Orbit. Phobos orbits Mars at a distance of 9,380 km from the center of the planet and has a period of 0.3189 days. Assu

Elenna [48]

Answer:

Explanation:

The relation between time period of moon in the orbit around a planet can be given by the following relation .

T² = 4 π² R³ / GM

G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .

Substituting the values in the equation

(.3189 x 24 x 60 x 60 s)² = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)

759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )

M = .06424 x 10²⁵

= 6.4 x 10²³ kg .

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3 years ago

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$