Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.
The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:
<em>70 km per second per megaparsec</em>.
We'll also need to know that 1 parsec = about 3.262 light years.
So the speed of your receding galaxy is
(Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =
(150 million) x (1 / 3,262,000) x (70 km/sec) =
<em>3,219 km/sec </em>in the direction away from us (rounded)
Answer:
0.25 L
Explanation:
= Initial pressure = 1 atm
= Initial Temperature = 20 °C
= Initial volume = 4.91 L
= Final pressure = 5.2 atm
= Final Temperature = -196 °C
= Final volume
From ideal gas law we have
The pressure experienced by the balloon is 0.25 L
Answer:
a) t = 1.47 h b) t = 1.32 h
Explanation:
a) In this problem the plane and the wind are in the same North-South direction, whereby the vector sum is reduced to the scalar sum (ordinary). Let's calculate the total speed
v = f -
v = 585 -32.1
v = 552.9 km / h
We use the speed ratio in uniform motion
v = x / t
t = x / v
t = 815 /552.9
t = 1.47 h
b) We repeat the calculation, but this time the wind is going in the direction of the plane
v= f -
v 585 + 32.1
v = 617.1 km / h
t = 815 /617.1
t = 1.32 h