Answer: The mass released from the top of the 60º incline.
Explanation:
The external forces acting upon any of the masses, are the same for both (not numerically): the gravity, the normal force and the friction force.
As the gravity is always directed downward, we can decompose it in a component normal to the surface (that is equal to the normal force as the body is not accelerating in this direction), and another which is parallel to the incline, and which is the responsible for the acceleration of the object along the incline (opposed by the friction force).
So, we can write the following:
Fx = mg sin θ - μk .Fn = m.g. sin θ - μk. mg cos θ= m a
We can easily see, that if the angle is larger, the acceleration component due to gravity will be larger, and at the same time, the friction force will be smaller, so the acceleration for the 60º will be the larger one.
Answer:
3.04 m/s
Explanation:
As the woman moves north, the raft moves south.
We then make the following assumptions for easier calculations
Let the motion of the woman = x
Let the motion of the raft = 4.2 - x
Using law of momentum, we see that
m*v = m1*v1, where
m = 56 kg
v = x
m1 = 149 kg
v1 = (4.2 -x)
Substituting the values into the equation, we have
56 * x = 149 (4.2 - x), opening the bracket
56x = 623 - 149x collecting like terms
205x = 623 Divide by 205
x = 623/205
x = 3.04 m/s
This then means that our answer is 3.04 m/s
Answer:
The gauge pressure is 1511.11 psi.
Explanation:
Given that,
Flow rate = 94 ft³/min
Diameter d₁=3.3 inch
Diameter d₂ = 5.2 inch
Pressure P₁= 15 psi
We need to calculate the pressure on other side
Using Bernoulli equation
![P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2](https://tex.z-dn.net/?f=P_%7B1%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_%7B1%7D%5E2%3DP_%7B2%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_%7B2%7D%5E2)
We know that,
![V=Av](https://tex.z-dn.net/?f=V%3DAv)
![v=\dfrac{V}{A}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7BV%7D%7BA%7D)
Where, V = volume
v = velocity
A = area
Put the value of v into the formula
![P_{1}+\dfrac{1}{2}\rho (\dfrac{V}{A_{1}})^2=P_{2}+\dfrac{1}{2}\rho (\dfrac{V}{A_{2}})^2](https://tex.z-dn.net/?f=P_%7B1%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20%28%5Cdfrac%7BV%7D%7BA_%7B1%7D%7D%29%5E2%3DP_%7B2%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20%28%5Cdfrac%7BV%7D%7BA_%7B2%7D%7D%29%5E2)
Put the value into the formula
![15+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2=P_{2}+\dfrac{1}{2}\times0.36\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2](https://tex.z-dn.net/?f=15%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes0.36%5Ctimes%28%5Cdfrac%7B2707.2%5Ctimes4%7D%7B%5Cpi%5Ctimes%283.3%29%5E2%7D%29%5E2%3DP_%7B2%7D%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes0.36%5Ctimes%28%5Cdfrac%7B2707.2%5Ctimes4%7D%7B%5Cpi%5Ctimes%285.2%29%5E2%7D%29%5E2)
![P_{2}=15+\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(3.3)^2})^2-\dfrac{1}{2}\times0.036\times(\dfrac{2707.2\times4}{\pi\times(5.2)^2})^2](https://tex.z-dn.net/?f=P_%7B2%7D%3D15%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes0.036%5Ctimes%28%5Cdfrac%7B2707.2%5Ctimes4%7D%7B%5Cpi%5Ctimes%283.3%29%5E2%7D%29%5E2-%5Cdfrac%7B1%7D%7B2%7D%5Ctimes0.036%5Ctimes%28%5Cdfrac%7B2707.2%5Ctimes4%7D%7B%5Cpi%5Ctimes%285.2%29%5E2%7D%29%5E2)
![P_{2}=1525.8\ psi](https://tex.z-dn.net/?f=P_%7B2%7D%3D1525.8%5C%20psi)
We need to calculate the gauge pressure
Using formula of gauge pressure
![P_{g}=P_{ab}-P_{atm}](https://tex.z-dn.net/?f=P_%7Bg%7D%3DP_%7Bab%7D-P_%7Batm%7D)
Put the value into the formula
![P_{g}=1525.8-14.69](https://tex.z-dn.net/?f=P_%7Bg%7D%3D1525.8-14.69)
![P_{g}=1511.11\ psi](https://tex.z-dn.net/?f=P_%7Bg%7D%3D1511.11%5C%20psi)
Hence, The gauge pressure is 1511.11 psi.