Answer:
(a) The magnitude of force is 116.6 lb, as exerted by the rod CD
(b) The reaction at A is (-72.7j-38.1k) lb and at B it is (37.5j) lb.
Explanation:
Step by step working is shown in the images attached herewith.
For this given system, the coordinates are the following:
A(0, 0, 0)
B(26, 0, 0)
And the value of angle alpha is 20.95°
Hope that answers the question, have a great day!
Answer:
![h_{max} = 51.8 cm](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%2051.8%20cm)
Explanation:
given data:
height of tank = 60cm
diameter of tank =40cm
accelration = 4 m/s2
suppose x- axis - direction of motion
z -axis - vertical direction
= water surface angle with horizontal surface
accelration in x direction
accelration in z direction
slope in xz plane is
![tan\theta = \frac{a_x}{g +a_z}](https://tex.z-dn.net/?f=%20tan%5Ctheta%20%3D%20%5Cfrac%7Ba_x%7D%7Bg%20%2Ba_z%7D)
![tan\theta = \frac{4}{9.81+0}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%20%5Cfrac%7B4%7D%7B9.81%2B0%7D)
![tan\theta =0.4077](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D0.4077)
the maximum height of water surface at mid of inclination is
![\Delta h = \frac{d}{2} tan\theta](https://tex.z-dn.net/?f=%5CDelta%20h%20%3D%20%5Cfrac%7Bd%7D%7B2%7D%20tan%5Ctheta)
![=\frac{0.4}{2}0.4077](https://tex.z-dn.net/?f=%20%3D%5Cfrac%7B0.4%7D%7B2%7D0.4077)
![\Delta h 0.082 cm](https://tex.z-dn.net/?f=%20%5CDelta%20h%20%200.082%20cm)
the maximu height of wwater to avoid spilling is
![h_{max} = h_{tank} -\Delta h](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%20h_%7Btank%7D%20-%5CDelta%20h)
= 60 - 8.2
![h_{max} = 51.8 cm](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%2051.8%20cm)
the height requird if no spill water is ![h_{max} = 51.8 cm](https://tex.z-dn.net/?f=h_%7Bmax%7D%20%3D%2051.8%20cm)
Answer:
b
Explanation:
i think do kill me if im wrong
Most Motor Vehicle catches in Florida in 2016 occurred on Motorcycles. Another option could be roads.
Answer:
![\Delta P=61,952.8\ lb/ft^2](https://tex.z-dn.net/?f=%5CDelta%20P%3D61%2C952.8%5C%20lb%2Fft%5E2)
Explanation:
Given
Airline flying at 34,000 ft.
Cabin pressurized to an altitude 8,000 ft.
We know that at standard condition ,density of air
![\rho =0.074\ lb/ft^3](https://tex.z-dn.net/?f=%5Crho%20%3D0.074%5C%20lb%2Fft%5E3)
We know that pressure difference
ΔP=ρ g ΔZ
Here ΔZ=34,000-8,000 ft
ΔZ=26,000 ft
![g= 32.2\ ft/s^2](https://tex.z-dn.net/?f=g%3D%2032.2%5C%20ft%2Fs%5E2)
ΔP=0.074 x 32.2 x 26,000
![\Delta P=61,952.8\ lb/ft^2](https://tex.z-dn.net/?f=%5CDelta%20P%3D61%2C952.8%5C%20lb%2Fft%5E2)
So pressure difference will be
.