To solve the problem, we enumerate all the given first. Then the required and lastly the solution.
Given:
V1= 1.56x10^3 L = 1560 L P2 = 44.1 kPa
P1 = 98.9 kPa
Required: V2
Solution:
Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,
PV = k
Applying to the problem, we have
P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L
<em>ANSWER: V2 = 3498.5 L</em>
<span><span>1.
</span>If the ramp has a length of 10 and has a
mechanical advantage (MA) of 5. Then we need to find the height of the ramp.
Formula:
MA = L / H
Since we already have the mechanical advantage and length, this time we need to
find the height .
MA 5 = 10 / h
h = 10 / 5
h = 2 meters
Therefore, the ramp has a length of 10 meters, a height of 2 meters with a
mechanical advantage of 5.</span>
Answer:
The angle is 
Explanation:
From the question we are told that
The distance of separation is 
The wavelength of light is 
Generally the condition for destructive interference is mathematically represented as
![dsin(\theta ) =[m + \frac{1}{2} ]\lambda](https://tex.z-dn.net/?f=dsin%28%5Ctheta%20%29%20%20%3D%5Bm%20%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%5D%5Clambda)
Here m is the order of maxima, first minimum (dark space) m = 0
So
![100 *10^{-6 } * sin(\theta ) =[0 + \frac{1}{2} ]600 *10^{-9}](https://tex.z-dn.net/?f=100%20%2A10%5E%7B-6%20%7D%20%2A%20%20sin%28%5Ctheta%20%29%20%20%3D%5B0%20%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%5D600%20%2A10%5E%7B-9%7D)
=> ![\theta = sin^{-1} [0.003]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20sin%5E%7B-1%7D%20%5B0.003%5D)
=> 
Answer:
V = 2.5 J/C
Explanation:
<u><em>Given:</em></u>
Energy = E = 20 J
Charge = Q = 8 C
<u><em>Required:</em></u>
Potential Difference = V = ?
<u><em>Formula:</em></u>
V = 
<u><em>Solution:</em></u>
V = 20/8
V = 2.5 J/C
The friction force is equal to the horizontal component of applied force 'p'. This horizontal component of force is pcosθ( θ = angle made by force with horizontal ).
Hence the frictional force is equal to pcosθ.