Answer:
Ve = 0.01625 [m³]
Explanation:
This type of problem can be solved by the following expression.

where_
Pi = Initial pressure = 2.5 x 10⁵ [Pa]
Vi = initial volume = 1.5 x 10⁻² [m³]
To = final temperature = 302 [K]
Po = final pressure = 1.2 x 10⁵ [Pa]
Vo = final volume [m³]
Ti = initial temperature = 302 [K]
Now the key to solving this problem is to make it clear that the temperature remains constant throughout the process. That is, it does not change.
![2.5*10^{5}*1.5*10^{-2} *302 = 1.2*10^{5}*V_{o}*302\\1132500=36240000*V_{o}\\V_{o}=0.03125 [m^{3} ]](https://tex.z-dn.net/?f=2.5%2A10%5E%7B5%7D%2A1.5%2A10%5E%7B-2%7D%20%2A302%20%3D%201.2%2A10%5E%7B5%7D%2AV_%7Bo%7D%2A302%5C%5C1132500%3D36240000%2AV_%7Bo%7D%5C%5CV_%7Bo%7D%3D0.03125%20%5Bm%5E%7B3%7D%20%5D)
But this calculated volume corresponds to the volume of the tank, now by means of the difference between the initial volume and the final volume we can calculate the volume of gas that escaped.
![V_{e}=0.03125-0.015\\V_{e}=0.01625 [m^{3} ]](https://tex.z-dn.net/?f=V_%7Be%7D%3D0.03125-0.015%5C%5CV_%7Be%7D%3D0.01625%20%5Bm%5E%7B3%7D%20%5D)
Answer:
μ = .4842
Explanation:
Assuming constant acceleration and the object is sliding on a level surface:
Let's start off by finding the acceleration of the object.
- v₀ = 19 m/s
- v = 0 m/s
- Δx = 38 m
- a = ?
v² = v₀² + 2aΔx
- 0² = 19² + 2a(38)
- 0 = 361 + 76a
- -361 = 76a
- a = -4.75 m/s²
Now, we can make a free-body diagram and see that the only force opposing the object's motion is the friction force -- this is the only force in the x-direction.
We can use Newton's 2nd Law: F = ma to solve for the coefficient of friction.
F = ma
- F = friction force = μ * N = μ * mg
Now, we have:
The mass cancels out and solving for the coefficient of friction gives us:
We know the acceleration is -4.75 m/s² and g = -9.81 m/s²:
- μ = -4.75/-9.81
- μ = .4842
The coefficient of kinetic friction between the surfaces, assuming constant acceleration and a flat surface, is 0.4842.
Well I belive it has something to do with kinetic energy and I think it depends on if ur going up or down hill
6300
N
Explanation:
We will employ the following equation of kinematics
x
=
1
2
a
t
2
which describes an object traveling in one dimension with a constant acceleration and an initial velocity of zero.
We know the displacement and the time elapsed so we may solve for the acceleration:
40.0
m
=
1
2
a
(
3.0
s
)
2
Solving for
a
yields:
a
=
8.89
m
s
2
Now, knowing the acceleration of the car as well as the mass of the car we can apply Newton's second law, which states
F
net
=
m
a
All we need to do is plug in our values for
m
and
a
:
F
net
=
710
kg
⋅
8.89
m
s
2
=
6311.9
N
Considering the fact that our final answer should have only two significant digits, we will round to the nearest hundred:
F
net
Answer:
around the ring of fire boundaries
Explanation: