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earnstyle [38]
2 years ago
13

QUICK SOMEONE PLEASE HELP!!!! I’LL MARK BRAINLIEST!!!

Physics
1 answer:
Brilliant_brown [7]2 years ago
8 0
Temperatures above 100 are gases
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coronavirus

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7 0
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When lamps with wattages greater than the rating of the luminaire are installed, a fire could occur because the luminaire is bei
sergejj [24]

Answer:

true

Explanation:

Yes, it is true.

As the wattage is more than the prescribed wattage, it becomes overheated.

6 0
3 years ago
Describe some ways that industry and agriculture use physical properties to separate substances.
musickatia [10]
This is more chemistry. But it is a process called fractional distillation, and it basically separates the long chained hydrocarbons from the short chained hydrocarbons through separation dependant on the boiling point of the crude oil.
3 0
2 years ago
Read 2 more answers
A. Group of cross country runners decided to go on an hour and a half run. During the first hour, they ran a total of 13 kilomet
pshichka [43]

Answer:

The average speed for the entire run is 12 km/h.

Explanation:      

The average speed is given by the following equation:

\overline{v} = \frac{d_{T}}{t_{T}}

Where:

d_{T}: is the total distance

t_{T}: is the total time

If during the first hour, they ran a total of 13 kilometers and then, they ran 5.0 kilometers during the next half an hour we have:

d_{T} = 13 km + 5 km = 18 km

t_{T} = 1 h + \frac{1}{2} h = 1.5 h

Hence, the average speed is:

\overline{v} = \frac{d_{T}}{t_{T}} = \frac{18 km}{1.5 h} = 12 km/h

Therefore, the average speed for the entire run is 12 km/h.

I hope it helps you!                                                                                      

3 0
3 years ago
The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

v_{rms} = \sqrt{ \frac{3RT}{M}}

So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

v_{rms,o2}^2 *M_{o2} =3RT_{o2}

solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

T_{o2} = \frac{(11.1*10^3)^2 *32.0*10^{-3}}{3*8.31}

T_{o2} =1.58*10^5 K

7 0
3 years ago
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