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2 years ago

Which of these experiments would make use of quantitative data?

2 answers:

Klio2033 [76]2 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]2 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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__________ and __________ both agreed that childhood experiences play an important role in the development of oneself.

oksano4ka [1.4K]

The following answers in the blank could be Sigmund Freud who found the psychoanalysis and Karen Horney who is known to be a psychoanalyst. Both of them has the idea that childhood experiences is necessary and plays the important role in the individual as he or she grows.

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2 years ago

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What renewable source of energy would be suited to chicago IL?

Kay [80]

Answer:

Wind is the primary renewable resource used for electric power generation in the state. In 2019, wind provided 97% of the state's renewable energy generation, and Illinois was sixth in the nation in utility-scale (1 megawatt or greater) wind capacity, with about 5,200 megawatts online.

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2 years ago

Identify which best describes the energy used to pluck guitar strings to make sound.

Novay_Z [31]

<span>The correct answer is: Mechanical Energy

Explanation:
As the guitar strings are plunked, the potential energy stored in the strings has an ability to make them vibrate. When the strings are vibrating, that potential energy is actually converted to the kinetic energy. Hence, the whole phenomena contains both the kinetic energy and the potential energy. The sum of kinetic energy and the potential energy is called Mechanical energy. Therefore, the correct answer is Mechanical Energy.</span>

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2 years ago

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A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball

lesantik [10]

The kinetic energy of 2.5 kg ball after collision is 27.09 J.

Answer:

Explanation:

In elastic collision, the sum of momentum of the objects before collision will be equal to the sum of momentum of the objects after collision.

We know that momentum is the product of mass and velocity acting on any object.

So, the conservation of energy in elastic collision leads to following equation:

$M_{1} u_{1} +M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

Since, the momentum is conserved ,the kinetic energy will also be conserved in elastic collision. So

$M_{1} u_{1} ^{2}+M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

Since initial velocity for M1 ball is zero, then

$M_{2} u_{2}=M_{1} v_{1}+M_{2} v_{2}$

and

$M_{2} u_{2} ^{2}=M_{1}v_{1} ^{2}+ M_{2}v_{2} ^{2}$

So, on solving all the above equation, we get an equation for velocity and that is

$\frac{2M_{2}u_{2} }{(M_{1}+M_{2} }$ =final velocity of ball with mass 2.5 kg

$v = \frac{2(5*3.5)}{2.5+5}=4.67 m/s$

So kinetic energy will be 1/2 mv2

Kinetic energy of 2.5 kg ball is $\frac{1}{2}*2.5*(4.67)^{2} =27.09 J$

So the kinetic energy of 2.5 kg ball after collision is 27.09 J.

6 0

3 years ago

A solenoid that is 78.8 cm long has a cross-sectional area of 15.9 cm2. There are 914 turns of wire carrying a current of 8.25 A

Harlamova29_29 [7]

Answer:

(a) Energy density will be equal to $57.31J/m^3$

(b) Total energy will be equal to 0.0718 J

Explanation:

It is given that length of solenoid l = 78.8 cm = 0.788 m

Cross sectional area $A=15.9cm^2=15.9\times 10^{-4}m^2$

Number of turns of the wire N = 914

Current in the solenoid i = 8.25 A

Inductance of the wire is equal to $L=\frac{\mu _0N^2A}{l}=\frac{4\pi \times 10^{-7}\times 914^2\times 15.9\times 10^{-4}}{0.788}=2.117\times 10^{-3}H$

(b) Total energy stored in magnetic field $U=\frac{1}{2}Li^2=\frac{1}{2}\times 2.11\times 10^{-3}\times 8.25^2=0.0718J$

(a) Energy density will be equal to

$U_b=\frac{0.0718}{15.9\times 10^{-4}\times 0.788}=57.31J/m^3$

7 0

2 years ago