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-BARSIC- [3]
3 years ago
13

Which of these experiments would make use of quantitative data?

Physics
2 answers:
Klio2033 [76]3 years ago
6 0
An experiment that would make use of quantitative data would be A. A study of the different amounts of time it takes water to evaporate completely.
denis23 [38]3 years ago
5 0

Answer;

A study of the different amounts of time it takes water to evaporate completely

Explanation;

-Quatitative data are anything that can be expressed as a number, or quantified. Examples of quantitative data are scores on achievement tests,number of hours of study, or weight of a subject. These data may be represented by ordinal, interval or ratio scales and lend themselves to most statistical manipulation.

-Qualitative data on the other hand, cannot be expressed as a number. Data that represent nominal scales such as gender, social economic status, religious preference are usually considered to be qualitative data.

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What is the strength of the electric field inside the membrane just before the action potential?
gtnhenbr [62]

Answer:

Incomplete question, check attachment for the graph needed to solve problem.

A 8.1nm........

Explanation:

Electric Field is given as

E=V/d

Where V is voltage

And d is the distance apart

E is the electric field

The voltage V just before action of potential is -70mV,

The value d=8.1nm

d=8.1×10^-9m

E=V/d

E=-70×10^-3/8.1×10^-9

E=-8.6×10^6 N/C

Then the magnitude of the electric field is 8.6×10^6N/C

5 0
3 years ago
Thermal Conductors don't have to be hot to transfer heat, explain a situation when a ice cube would still transfer heat to anoth
barxatty [35]
An ice cube would transfer heat to another object whose temperature
is lower than zero°C (32°F).

A block of "dry ice" is sitting there at a temperature of -78°C (-109°F).
An ice cube helps to melt dry ice nice and fast.

If you could find a block of solid nitrogen, its temperature would be
63K (-210°C, -346°F).  An ice cube would transfer heat to that baby
so fast that it would instantly boil.

6 0
3 years ago
Which statement describes steps involved in the production of hydroelectric power?
viktelen [127]

Answer:

Answer is D.......Falling water turns a turbine that helps generate electricity.

Explanation:

Hydropower plants capture the energy of falling water to generate electricity. A turbine converts the kinetic energy of falling water into mechanical energy. Then a generator converts the mechanical energy from the turbine into electrical energy.

4 0
3 years ago
Read 2 more answers
With which of the following statements would Whorf and Sapir agree?
ASHA 777 [7]
Hi!

The answer is <span>B. Language influences how people understand their world.

Hope this helps!

-Payshence xoxo</span>
6 0
2 years ago
Read 2 more answers
Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit
Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c)A=0.01297m^2

Explanation:

A)

The kinetic energy is defined as:

\frac{m*vel^2}{2} (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ (\frac{mvel^2}{2})=\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}

Δ (\frac{mvel^2}{2})=\frac{m}{2}*(vel_2^2-vel_1^2)

Where 1 and 2 subscripts mean initial and final state respectively.

Δ(\frac{mvel^2}{2})=\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)

We already know the last quantity: \frac{m}{2} *(vel_1^2-vel_2^2)=-Δ (\frac{mvel^2}{2})=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}

The exit state is a liquid-vapor mixture, so its enthalpy is:

h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}

Finally, the work can be obtained:

W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW

C) For the area, consider the equation of mass flow:

m=p*vel*A where p is the density, and A the area. The density is the inverse of the specific volume, so m=\frac{vel*A}{v}

The specific volume of the inlet steam can be read also from the steam tables, and its value is: 0.08643\frac{m^3}{kg}, so:

A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2

Download pdf
7 0
3 years ago
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