Less gas will be collected because some of the gases will escape from the open cylinder valve.
Cylinders used to store carbon dioxide will have thicker walls than those of butane because of higher pressures.
<h3>What are compressed gases?</h3>
Compressed gases are gases which are compressed under high pressure in gas cylinders.
Cylinder valves are used to reduce the pressure of the compressed gases and in the process, some of the gas molecules escape.
Since the cylinder valve is open and the gas is collected at atmospheric pressure, less gas will be collected because some of the gases will escape.
Since, the carbon dioxide not liquefy under pressure compared to butane, the cylinders used to store carbon dioxide will have thicker walls than those of butane.
Learn more about compressed gases at: brainly.com/question/518065
The gravitational force between two objects is given by:
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation
In this problem, the first object has a mass of
, while the second "object" is the Earth, with mass
. The distance of the object from the Earth's center is
; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
Answer:
0.61°
Explanation:
Since the box move at constant velocity, it means there is no acceleration then we can say it has a balanced force system.
Pulling force= resistance force
From the formula for pulling force,
F(x)= Fcos(θ)
= 425×cos(35.2)
=347N
The force exerted downward at an angle of 35.2° below the horizontal= Fsin(θ)= 425sin(35.2)
=425×0.567=245N
Resistance force= (325N+ 245N) (α)= 570N(α)
We can now equates the pulling force to resistance force
570 (α)= 347N
(α)= 347/570
= 0.61
Answer:
Explanation:
If the dragster attains the speed equal to that of the car which is moving with constant velocity of v₀ , before the two close in contact with each othe , there will not be collision .
So the dragster starting from rest , must attain the velocity v₀ in the maximum time given that is tmax .
v = u + a t
v₀ = 0 + a tmax
tmax = v₀ / a
The value of tmax is v₀ / a .
