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2 years ago

A student heats a liquid on a burner. What happens to the portion of liquid that first begins to warm? What all applys

2 answers:

5 0

For water, as the temperature goes up the density of the liquid water decreases therefore it becomes less dense so it tends to rise upward. And since the temperature here is higher, heat flows from the higher temperature to the lower so it releases energy to the environment. Therefore, the correct statements are 2, 3 and 5.

sineoko [7]2 years ago

5 0

Answer:

It becomes less dense.

It absorbs energy from the environment.

It rises upward.

Explanation:

Took the test!

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In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to

dexar [7]

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, $a=\dfrac{v-u}{t}$ ....(1)

The third equation of kinematics is as follows :

$v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s$

Hence, it will take 25 seconds to increase the speed.

6 0

2 years ago

A car is traveling in a race.The car went from initial velocity of 35m/s to the final velocity of 65m/s in 5 seconds what was th

I am Lyosha [343]

Acceleration is the change in velocity divided by time. The change in velocity is -30m/s and time is 5s. If you divide -30m/s by 5s, you get -6m/s².

8 0

2 years ago

Read 2 more answers

A cement factory emits 900 kilograms of CO2 to produce 1,000 kilograms of cement. A fully grown tree removes six kilograms of CO

Dmitriy789 [7]

<h3>Answer;</h3>

To make the factory carbon neutral, the owners need to grow 15000 trees over an area of land measuring 75 acres

<h3>Explanation and solution;</h3>

From the information;

900 kg CO2 = 1000 kg Cement

1 tree = 6 kg CO2

1 acre = 200 trees

100000 kg Cement will require;

=(900 × 100000)/1000

= 90,000 kg of CO2

But 1 tree = 6 kg of CO2

Number of trees = 90,000/6

= 15,000 trees

But, 1 acre = 200 trees

Number of acres = 15,000/200

= 75 acres of land

Therefore;

To make the factory carbon neutral, the owners need to grow 15000 trees over an area of land measuring 75 acres

6 0

2 years ago

Read 2 more answers

You hang different masses M from the lower end of a vertical spring and measure the period T for each value of M. You use Excel

Svetradugi [14.3K]

Answer:

a)693.821N/m

b)17.5g

Explanation:

We the Period T we can find the constant k,

That is

$T = 2 \pi \sqrt{\frac{m}{k}}$

squaring on both sides,

$T^2=\frac{4\pi^2}{k}M +\frac{4\pi^2}{k}m_{spring}$

where,

M=hanging mass, m = spring mass,

k =spring constant

T =time period

a) So for the equation we can compare, that is,

$y=T^2=0.0569x+0.0010$

the hanging mass M is x here, so comparing the equation we know that

$\frac{4\pi^2}{k}=0.0569\\k= \frac{4\pi^2}{0.0569}\\k=693.821N/m$

b) In order to find the mass of the spring we make similar process, so comparing,

$\frac{4\pi^2}{k}m =0.001\\m=\frac{0.004k}{4\pi^2} =\frac{0.001*693.821}{4\pi^2}\\m=0.0175kg\\m=17.5g$

3 0

2 years ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$