V₁(O2) = 6.50<span> L
</span>p₁(O2) = 155 atm
V₂(acetylene) = <span>4.50 L
</span>p₂(acetylene) =?
According to Boyle–Mariotte law (At constant temperature and unchanged amount of gas, the product of pressure and volume is constant) we can compare two gases that have ideal behavior and the law can be usefully expressed as:
V₁/p₁ = V₂/p₂
6.5/155 = 4.5/p₂
0.042 x p₂ = 4.5
p₂ = 107.3 atm
I'm guessing that you mean like this:
-- The ruler is held with zero at the bottom, and the centimeter markings
increase as you go up the ruler.
-- You place your fingers with the ruler and the zero mark between them.
-- The number where you catch the ruler is the distance it has fallen.
Then, all we have to find is the time it takes for the ruler to fall 11.3 cm .
Here's the formula for the distance an object falls from rest
in a certain time:
Distance = (1/2) (gravity) (time)²
On Earth, the acceleration of gravity is 9.8 m/s².
So we can write ...
11.2 cm = (1/2) (9.8 m/s²) (time)²
or
0.112 meter = (4.9 m/s²) (time)²
Divide each side
by 4.9 m/s² : (0.112 m) / (4.9 m/s²) = time²
(0.112 / 4.9) sec² = time²
Square root
each side: time = √(0.112/4.9 sec²)
= √ 0.5488 sec²
= 0.74 second (rounded)
Answer:
a) 69.3 m/s
b) 18.84 s
Explanation:
Let the initial velocity = u
The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively
uᵧ = u sin 40° = 0.6428 u
uₓ = u cos 40° = 0.766 u
We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m
The range of a projectile motion is given as
R = uₓt
where t = total time of flight
1000 = 0.766 ut
ut = 1305.5
The vertical distance travelled by the projectile rocks,
y = uᵧ t - (1/2)gt²
y = - 900 m (900 m below the crater's level)
-900 = 0.6428 ut - 4.9t²
Recall, ut = 1305.5
-900 = 0.6428(1305.5) - 4.9 t²
4.9t² = 839.1754 + 900
4.9t² = 1739.1754
t = 18.84 s
Recall again, ut = 1305.5
u = 1305.5/18.84 = 69.3 m/s
It is a mechanical wave and cannot travel through a vacuum.
