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Molodets [167]
3 years ago
6

In the forest (phrase—noun, but no subject and no verb) Smashing against the bat (phrase—verb but no subject) Before our first c

lass (phrase) After the snowstorm (phrase) I am aware of the error (clause—subject and verb) When the saints go marching in (clause) Between a rock and a hard place (phrase) Broken shards of glass (phrase) Because of her beautiful smile (phrase) Because she smiled at him (clause)
Physics
1 answer:
gayaneshka [121]3 years ago
6 0

Yes all those are correct, but I don't know what the question was

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Heat from burning fuel warms the walls of the firebox section of the furnace in
nydimaria [60]

Heat from burning fuel warms the walls of the firebox section of the furnace in

A. a hot-water heating system.

B. a hot-air heating system.

C. a compressor compartment.

D. an evaporation system.

7 0
2 years ago
The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

6 0
3 years ago
A hydrogen atom that has an electron in the n = 2 state absorbs a photon. What wavelength must the photon possess to send the el
Deffense [45]

Answer:

486nm

Explanation:

in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that

\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\

Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state

4 0
2 years ago
Which compound is most likely a covalent compound?<br> Melting Boiling
mote1985 [20]

Answer:

compound A

Explanation:

because covalent compounds can't conduct electricity and because they also have low boiling points

7 0
3 years ago
Choose ALL of the answers that would support the theory of Continental Drift?
saveliy_v [14]
The answers are B, D, E!
7 0
2 years ago
Read 2 more answers
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