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2 years ago

What is one advantage of using primary sources when doing research on an

1 answer:

5 0

One advantage of using primary sources when doing research on an experiment is, it contain more details about the experiment. Thus, option D is right .

To find the answer, we have to know more about the Primary sources.

<h3>What is the advantages of primary sources?</h3>

Primary sources are the direct or first-hand excerpts that scientists have independently written based on their experiments in order to support their research.
Utilizing original sources encourages analytical and critical thought in relation to the research.

It aids in exploring from various angles, which leads to the discovery of extra facts.
Primary data aids in navigating the conflicts. Since it is a direct resource related to the experiment, it serves as evidence for the data.

Thus, we can conclude that, Primary sources contain more details about the experiment.

Learn more about the Primary sources here:

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Would you classify petroleum as a renewable or nonrenewable resource? Justify your answer in two or more complete sentences.

olchik [2.2K]

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3 years ago

How to be good at rocket league

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3 years ago

Read 2 more answers

You need to estimate the floor area of your unusual oval shaped living room from memory you remember that it is longer than wide

REY [17]

Answer:

Our estimate not done correctly because the width was unrealistic.

Explanation:

Given that,

Length = 6 m

Width = 0.5 m

We need to calculate the area of rectangular floor

Using formula of area

$A=l\times w$

Where, l = length

w= width

Put the value into the formula

$A=6\times0.5$

$A=3\ m^2$

We need to find why this estimate not done correct

Using given data

We can say that,

We assume width was unrealistic.

The length of room is 6 m.

We know that,

$1\ m=3.28\ feet$

So, the length would be 20 feet.

The width of room is 0.5 m.

The width of room would be 1.64 feet.

So, the area will be

$A=20\times1.64$

$A=32.8\ feet^2$

Hence, Our estimate not done correctly because the width was unrealistic.

4 0

3 years ago

An example of intentionally increasing friction?

eimsori [14]

You can increase friction intentionally by rubbing two objects together.

5 0

3 years ago

Read 2 more answers

A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i

AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume $v$ and enthalpy $h$ as,

$h_1 = 95.96Btu/lb$ ( $h_1 = h_f$ at $2psia$ )

$v_1 = 0.016238ft^3/lb$ ( $v_1 = v_f$ at $2psia$ )

$w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb$

$w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb$

$h_3 = 1364.0Btu/lb$

$s_3 = 1.5073Btu/lb.R$

( at $P_3 = 1500psia$ & $T_3 = 800^0F$ )

$P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765$

( $S_f$ & $S_{fg}$ when pressure is 2psia)

$h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb$

$n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb$

Therefore,

$q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb$

To calculate the mass flow rate of steam in the cycle, we use the formula

$W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s$

where $1Kj = 0.947817 Btu$

The power output and the rate of heat addition are calculated thus,

$W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW$

$Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s$

The thermal efficiency of the cycle can be found thus;

$n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343$

= 34.3%

5 0

3 years ago