Answer:
b) Betelgeuse would be 
times brighter than Sirius
c) Since Betelgeuse brightness from Earth compared to the Sun is 
the statement saying that it would be like a second Sun is incorrect
Explanation:
The start brightness is related to it luminosity thought the following equation:
(1)
where 
is the brightness, 
is the star luminosity and 
, the distance from the star to the point where the brightness is calculated (measured). Thus:
b) 
and 
where 
is the Sun luminosity (
) but we don't need to know this value for solving the problem. 
is light years.
Finding the ratio between the two brightness we get:
c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 
. Then
Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.
The prefix for oxygen in As2O5 is PENTA.
Penta is used to show that the number of oxygen atoms present in the compound is five. Penta means five, while mono mean one, di means two, tri means three and tetra mans four and so on. The chemical name for the given compound is Arsenic pentoxide.
<span>Velocity tells you what speed a moving object travels at and in what direction.</span>
Answer:
The metabolic power for starting flight=134.8W/kg
Explanation:
We are given that
Mass of starling, m=89 g=89/1000=0.089 kg
1 kg=1000 g
Power, P=12 W
Speed, v=11 m/s
We have to find the metabolic power for starting flight.
We know that
Metabolic power for starting flight=
Using the formula
Metabolic power for starting flight=
Metabolic power for starting flight=134.8W/kg
Hence, the metabolic power for starting flight=134.8W/kg
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
