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katrin [286]
3 years ago
6

Use the information to answer the following question.

Physics
1 answer:
Nostrana [21]3 years ago
5 0

Answer:

The answer is B. :)

Explanation:

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\nljbvjlsfhglskhg;skgh;sfjs;ojsr;ogjs;dofhs;dkhs;lifhns
vladimir1956 [14]
That is nonsense, also known as jibberish
8 0
3 years ago
ou are to drive to an interview in another town, at a distance of 300 km on an expressway. The interview is at 11:15 a.m. You pl
Shkiper50 [21]

Answer:

133.62 kmh.

Explanation:

Time provided = 3.25 hours.

Distance to be covered 300 km

Times spent in first  100 km = 1 hour

Time spent in next 43 km

= 43 / 40 = 1.075 hours

Total time spent = 2.075 hours

Total distance covered = 143 km

Distance remaining = 300 - 143

=157 km .

Time remaining = 3.25 - 2.075

= 1.175

Speed required = Distance remaining / time remaining

= 157 / 1.175

= 133.62 kmh.

8 0
3 years ago
Hi people; there's a question that has been so confusing for me and a friend of mine.
Alborosie

Answer:

The correct answer is: 0°C + 0°C = 32°F

Explanation:

We need to remember the conversion equation from Celsius to Fahrenheit:

y^{\circ}F=(x^{\circ}C * \frac{9}{5})+32

In our case x = 0, then y will be:

(0^{\circ}C * \frac{9}{5})+32=32

y=32^{\circ}F

Now 0°C + 0°C is just 0°C because if we add a body at a certain temperature to another body with the same temperature the total temperature will the same.

Then, knowing that 0°C = 32°F we can conclude that:

0^{\circ}C+0^{\circ}C=32^{\circ}F

I hope it helps you!              

7 0
3 years ago
What is one reason why it is very difficult to directly take a picture of an extrasolar planet?
Setler [38]

Answer:

Extrasolar planets are very dim light sources compared to their stars. At visible wavelengths, they generally have less than a millionth of the brightness of their parent star. It is extremely difficult to detect this type of dim light source, and in addition, the parent star has dazzling light that almost makes it impossible.

5 0
3 years ago
Read 2 more answers
What is the energy of the photon that could cause (a) an electronic transition from then= 4 state to the n 5 state of hydrogen a
Fiesta28 [93]

Answer:

a) E_photon =0.306 eV

b) E_photon =0.166 eV

Explanation:

The energy of the photon (E) for n^th orbit of the hydrogen atom is given as:

E_photon = E_o(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})

where,

E_o = 13.6 eV

n = orbit

a) Now for the transition from n = 4 to n = 5

E_photon =13.6(\frac{1}{4^{2}}-\frac{1}{5^{2}})

E_photon =0.306 eV

b) Now for the transition from n = 5 to n = 6

E_photon =13.6(\frac{1}{5^{2}}-\frac{1}{6^{2}})

E_photon =0.166 eV

7 0
3 years ago
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