Star 1 - 4 hours right ascension
Star 2 - 3 hours right ascension
Subtracting hours right ascension
4 hours right ascension - 3 hours right ascension = 1 hours right ascension.
Thus,
star 1 will rise 1 hour before star 2
Answer:
78 percent
Explanation:
I guess that's the right answer
Answer:
Boyle's Law

Explanation:
Given that:
<u><em>initially:</em></u>
pressure of gas, 
volume of gas, 
<em><u>finally:</u></em>
pressure of gas, 
volume of gas, 
<u>To solve for final volume</u>
<em>According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.</em>
<em>According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.</em>
But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.
Here nothing is said about the temperature, so we consider the Boyle's Law which states that <em>at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.</em>
Mathematically:



Answer:
Explanation:
Given that,
Mass of ball m = 2kg
Ball traveling a radius of r1= 1m.
Speed of ball is Vb = 2m/s
Attached cord pulled down at a speed of Vr = 0.5m/s
Final speed V = 4m/s
Let find the transverse component of the final speed using
V² = Vr²+ Vθ²
4² = 0.5² + Vθ²
Vθ² = 4²—0.5²
Vθ² = 15.75
Vθ =√15.75
Vθ = 3.97 m/s.
Using the conservation of angular momentum,
(HA)1 = (HA)2
Mb • Vb • r1 = Mb • Vθ • r2
Mb cancels out
Vb • r1 = Vθ • r2
2 × 1 = 3.97 × r2
r2 = 2/3.97
r2 = 0.504m
The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m
The required time,
Using equation of motion
V = ∆r/t
Then,
t = ∆r/Vr
t = (r1—r2) / Vr
t = (1—0.504) / 0.5
t = 0.496/0.5
t = 0.992 second