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Eddi Din [679]
3 years ago
6

The star Antares has an apparent magnitude of 1.0, whereas the star Procyon has an apparent magnitude of 0.4. Which star appears

brighter in the sky?
Physics
1 answer:
Anika [276]3 years ago
6 0

Answer:

Procyon appears brighter in the sky

Explanation:

Apparent magnitude of star Antares = 1.0

Apparent magnitude of star Procyon = 0.4

Pogson's Ratio

m₂-m₁ =  -2.50 log(B₂/B₁)

where, m is the apparent magnitude

B = Brightness of star or flux coming towards us (W/m²)

∴Larger magnitudes correspond to fainter stars so here 1>0.4 which means Antares appears dimmer and Procyon appears brighter.

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Answer:

Accuracy measures how close results are to the true or known value. Precision, on the other hand, measures how close results are to one another.

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The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o
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Answer:

F_f=840N

Explanation:

From the question we are told that

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A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin
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966 mph

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6 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

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First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

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W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

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2.537 ---> 20%

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So the student is consuming 12.685 KCals each time he runs up the stairs.

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1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

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Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

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