Say we have a cylinder
that has a height of dx, we see that the cylinder has a volume of: <span>
<span>Vcylinder = πr^2*h = π(5)^2(dx) = 25π dx
Then, the weight of oil in this cylinder is:
Fcylinder = 50 * Vcylinder = (50)(25π dx) = 1250π dx.
Then, since the oil x feet from the top of the tank needs to
travel x feet to get the top, we have:
Wcylinder = Force x Distance = (1250π dx)(x) = 1250π x dx.
<span>Integrating from x1 to x2 ft gives the total work to be: (x1
= distance from top liquid level to ground level; x2 = distance from bottom
liquid level to ground level)</span>
<span>W = ∫ 1250π x dx
<span>W = 1250π ∫ x dx
W = 625π * (x2 – x1)</span></span></span></span>
<span>x2 = 14 ft + 15 ft = 29 ft</span>
x1 = 14 ft + 1 ft = 15
ft
<span>
W = 625π * (29^2 - 15^2)
<span>W = 385,000π ft-lbs
= 1,209,513.17 ft-lbs</span></span>
Answer:

Explanation:
Please find the image for the question as attached file.
Solution -
Given -
First of all we will calculate the velocity at point C,
As per newton's third law of motion-

Substituting the given values in above equation, we get -

Now we will determine the radius of curvature for the curve shown in the attached image

Differentiating on both the sides, we get -
meter
Acceleration on curved path

Final acceleration

Answer:
t = 0.55[sg]; v = 0.9[m/s]
Explanation:
In order to solve this problem we must establish the initial conditions with which we can work.
y = initial elevation = - 1.5 [m]
x = landing distance = 0.5 [m]
We set "y" with a negative value, as this height is below the table level.
in the following equation (vy)o is equal to zero because there is no velocity in the y component.
therefore:
![y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7Bo%7D%2At%20-%20%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%5C%5C%20%20%20where%3A%5C%5C%28v_%7By%7D%29_%7Bo%7D%3D0%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%5Bsg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%5D%5C%5C%20-1.5%20%3D%200%2At%20-4.905%2At%5E%7B2%7D%20%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B1.5%7D%7B4.905%7D%20%7D%20%5C%5Ct%3D0.55%5Bs%5D)
Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.
![(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7B0.5%7D%7B0.55%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D0.9%5Bm%2Fs%5D)
Answer:
the answer would be B your welcome
Explanation:
Answer:
d. ovaries, uterus, and Fallopian tubes
Explanation: