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Leona [35]
3 years ago
8

__________ research is designed to measure the association between variables that are not manipulated by the researcher.

Physics
1 answer:
Sliva [168]3 years ago
3 0

Answer:

Option c. Correlational

Explanation:

Correlational research is basically a method of carrying out a research where an experiment can not be conducted and where a researcher has to figure out that whether the two variable are in association with each other and if these are related then in what way the association between these two exists without any manipulation from the researcher's end.

It seeks to understand, measure and make assessment of the relationship or  

association between these variables but can not find out if a variable is caused by the other variable.

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] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I
Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

\mathtt{=( 0.42\times 9000\times 1055.06) J}

= 3988126.8  J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec =\dfrac{750}{3.99}

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned  = \dfrac{1.1}{100} \times 187.97 \  lb/s

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0
2 years ago
A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect
zaharov [31]

Answer:

X=X_o+\dfrac{1}{2}gt^2

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

\dfrac{1}{2}mV^2=mg(X-X_o)

So

V=\sqrt{2g(X-X_o)}

We know that

\dfrac{dX}{dt}=V

\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}

\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX

At t= 0 ,X=Xo

So we can say that

X=X_o+\dfrac{1}{2}gt^2

So the length of cable after time t

X=X_o+\dfrac{1}{2}gt^2

6 0
3 years ago
You observe three carts moving to the left. Cart A moves to the left at nearly constant speed. Cart B moves to the left, gradual
Lady bird [3.3K]

Answer:cart B

Explanation:

For cart A speed is constant therefore there is no acceleration because acceleration is rate of change of velocity

thus there is no net force

For cart B there is change in velocity in the left direction , so there is net acceleration towards left

Force=mass\times acceleration

so there is net force in the left direction

For cart C there is decrease in velocity i.e. negative acceleration or deceleration . Therefore there is a net force towards right which opposes the motion                

6 0
3 years ago
An organism’s scientific name consists of a. its class name and its family name. b. its kingdom name and its phylum name. c. its
mr Goodwill [35]
An organism scientific name consist of : C. its genus name and its species name
The first part of the name is taken from the Genus and the second part of the name is taken from the species
hope this helps

7 0
3 years ago
A brick falls to the ground. if the time for the collision of the brick and the ground is increased by a factor of 4, the force
melomori [17]

Answer:

By a factor of 1/4.

Explanation:

The impulse force that applies to an object undergoing rapid deceleration just before coming to a stop on the ground is given by the following formula,

\\\begin{aligned}\\\small F &=\small \frac{\Delta (mV)}{\Delta T}\end{aligned}

in which \small \Delta (mV) , \small \Delta t represent the change in momentum and the time taken for that change.

If one increases the time that is taken for the momentum change (which remains constant for this situation) by a factor 4 and if that new force is represented by \small F_1, the following manipulation confirms the answer to this question.

\begin{aligned}\\\small F_1 &=\small \frac{\Delta (mV)}{4\Delta t}\\\\&=\small \frac{1}{4}\times\bigg[\frac{\Delta (mV)}{\Delta t}\bigg]\\\\&=\small \frac{1}{4}F\end{aligned}

Here \small F is the force that was applied to the object previously.

#SPJ4

4 0
2 years ago
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