<span>A change in the pressure of a gas results in a more significant change in volume than it would in a liquid.  is the statement that accurately describes the property of gas. Gas only depends on how you store it. the bigger the space the wider gas can expand, the smaller the space, the more compress the gas can become.</span>
        
                    
             
        
        
        
Using the equation v(average)=x traveled/time
v = 100/2.5
You get 40 kilometers per hour
Hope this helped!
        
             
        
        
        
Answer:
398.3 m, 334.2 m
Explanation:
The magnitude of the displacement vector is
v = 520 m
And its direction is
 
 
measured as north of east.
The x-component of this vector is given by:

While the y-component is given by

 
        
             
        
        
        
Answer:
J. Robert Oppenheimer
Explanation:
He led the Manhattan project and created the first nuclear bomb in WWII
 
        
             
        
        
        
Refer to the figure shown below, which is based on the given figure.
d = the horizontal distance that the projectile travels.
h = the vertical distance that the projectile travels.
Part A
From the geometry, obtain
d = X cos(α)                     (1a)
h = X sin(α)                      (1b)
The vertical and horizontal components of the launch velocity are respectively
v = v₀ sin(θ - α)               (2a)
u = v₀ cos(θ - α)             (2b)
If the time of flight is t, then
vt - 0.5gt² = -h
or
0.5gt² - vt - h = 0             (3a)
ut = d                                (3b)
Substitute (1a), (1b), (2a), (2b) (3b) into (3a) to obtain

![4.9[ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha }  ]^{2} - v_{0} sin(\theta -  \alpha ) [ \frac{X cos \alpha }{v_{0} cos(\theta -  \alpha } ] - X sin \alpha  = 0](https://tex.z-dn.net/?f=4.9%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%20%5D%5E%7B2%7D%20-%20v_%7B0%7D%20sin%28%5Ctheta%20-%20%20%5Calpha%20%29%20%5B%20%5Cfrac%7BX%20cos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%7D%20%5D%20-%20X%20sin%20%5Calpha%20%20%3D%200)
Hence obtain
![aX^{2}-bX=0 \\ where \\ a=4.9[ \frac{cos \alpha }{v_{0} cos(\theta -  \alpha )}]^{2} \\  b = cos \alpha \,  tan(\theta -  \alpha ) + sin \alpha](https://tex.z-dn.net/?f=aX%5E%7B2%7D-bX%3D0%20%5C%5C%20where%20%5C%5C%20a%3D4.9%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%5D%5E%7B2%7D%20%5C%5C%20%20b%20%3D%20cos%20%5Calpha%20%5C%2C%20%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%20%2B%20sin%20%5Calpha%20)
The non-triial solution for X is

Answer:
![X= \frac{sin \alpha  + cos \alpha  \, tan(\theta -  \alpha )}{4.9 [ \frac{cos \alpha }{v_{0} \, cos(\theta -  \alpha )}  ]^{2}}](https://tex.z-dn.net/?f=X%3D%20%5Cfrac%7Bsin%20%5Calpha%20%20%2B%20cos%20%5Calpha%20%20%5C%2C%20tan%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%7B4.9%20%5B%20%5Cfrac%7Bcos%20%5Calpha%20%7D%7Bv_%7B0%7D%20%5C%2C%20cos%28%5Ctheta%20-%20%20%5Calpha%20%29%7D%20%20%5D%5E%7B2%7D%7D%20)
Part B
v₀ = 20 m/s
θ = 53°
α = 36°
sinα + cosα tan(θ-α) = 0.8351
cosα/[v₀ cos(θ-α)] = 0.0423
X = 0.8351/(4.9*0.0423²) = 101.46 m
Answer:  X = 101.5 m