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ongnoine Newbie
ongnoine
ongnoine

Joined: 01 March 2012
Posts: 5

Posted: 02 March 2012 at 3:52pm | IP Logged
Question 1/ A tank is full of water. Find the work required to pump the water out of the spout. (Use 9.8 m/s2 for g. Use 1000 kg/m3 as the density of water. Assume r = 9 m and h = 3 m.


2/A tank is is half full of oil that has a density of 900 kg/m3. Find the work W required to pump the oil out of the spout. (Use 9.8 m/s2 for g. Assume r = 12 m and h = 4 m


akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5275

Posted: 02 March 2012 at 6:49pm | IP Logged
Originally posted by ongnoine

A tank is full of water. Find the work required to pump the water out of the spout. Use the fact that water weighs 62.5 lb/ft3. (Assume r = 6 ft, R = 12 ft, and h = 15 ft.)


Let the origin be at the center of the bottom of the tank. Consider a slice of height dx at a height x above the origin.
Let y = the radius of the slice

y = m x + r,
where m = (R-r)/h

Volume of the slice
= PI * y^2 dx
= PI * (m x + r)^2 dx
Weight of water in the slice
= d * PI * (m x + r)^2 dx (where d = density)

The water in the slice has to be lifted by height h-x
The work done in lifting the water in the slice
= d * PI * (m x + r)^2 * (h-x) dx

Total work done
W = integral(0,h) d * PI * (m x + r)^2 * (h-x) dx
W = PI d integral(0,h) (m x + r)^2 * (h-x) dx
W = PI d integral(0,h) (m^2 x^2 + 2mxr + r^2) * (h-x) dx
W = PI d integral(0,h) (m^2 x^2 h - m^2 x^3 + 2mxrh - 2mx^2r + r^2 h - r^2 x) dx
W = PI d (m^2 h^4 /3 - m^2 h^4 /4 + 2 m r h^3 /2 - 2 m r h^3 /3 + r^2 h^2 - r^2 h^2/2)
W = PI d (m^2 h^4 /12 + m r h^3 /3 + r^2 h^2 /2)

Substitute m = (R - r)/h = (12 - 6)/15 = 0.4 and the values of other variables

W = 3.14 * 62.5 (0.4^2 15^4 /12 + 0.4 * 6 * 15^3 /3 + 6^2 15^2 /2)
W = 4.66 * 10^7 lb-ft (Answer)

Check the calculations
ongnoine Newbie
ongnoine
ongnoine

Joined: 01 March 2012
Posts: 5

Posted: 02 March 2012 at 6:52pm | IP Logged

Wow...thanks you so much!!! I did all of them...so you dont need to solve the other 2 ...Thank for helping me!

Miggi Groupbie
Miggi
Miggi

Joined: 02 June 2006
Posts: 142

Posted: 04 March 2012 at 7:53pm | IP Logged
Light rays from the Sun, which is at an angle of 35 degrees above the western horizon, strike the still surface of a pond. What is the angle of reflection of the rays that leave the pond surface?
 
 
Sunlight strikes the surface of a lake at an angle of incidence of30.0. At what angle with respect tothe normal would a fish see the Sun?
 
I am getting 22 degree but not sure if its right.
 
 
A converging lens and a diverging lens, separated by a distance of 30 cm, are used in combination. The converging lens has a focal length of 15 cm. The diverging lens is of unknown focal length. An object is placed 60 cm on left of the converging lens; the final image  is formed 6cm  before the diverging lens. What is the focal length of the diverging lens?



Edited by Miggi - 04 March 2012 at 9:16pm
Miggi Groupbie
Miggi
Miggi

Joined: 02 June 2006
Posts: 142

Posted: 04 March 2012 at 9:07pm | IP Logged
Q the angle of incidence of a ray of light in air is adjusted gradually as it enters a shallow tank made of plexigals and filled with carbon disulfide. The tank is open at the top -the light ray passes directly  from air into the carbon disulfide. Is there a minimum angle of incidence for which light is transmitted into the carbon disulfide but not into te plexigals at the bottom of the tank? If so find the angle. If not explain why not? Index of refraction values are air 1.00, plexigals 1.51, carbon disulfide 1.628. 
akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5275

Posted: 05 March 2012 at 1:40am | IP Logged
Originally posted by Miggi

Light rays from the Sun, which is at an angle of 35 degrees above the western horizon, strike the still surface of a pond. What is the angle of reflection of the rays that leave the pond surface?
Angle of incidence = 90 deg - 35 deg = 55 deg
Angle of reflection = angle of incidence = 55 deg


Sunlight strikes the surface of a lake at an angle of incidence of30.0. At what angle with respect tothe normal would a fish see the Sun?
Angle of incidence i = 30.0 deg
Let r = angle of refraction
sin i/sin r = n
r = sin-1(sin i/n)
r = sin-1 (sin 30.0 deg / 1.33)
r = 22.1 deg
 
A converging lens and a diverging lens, separated by a distance of 30 cm, are used in combination. The converging lens has a focal length of 15 cm. The diverging lens is of unknown focal length. An object is placed 60 cm on left of the converging lens; the final image  is formed 6cm  before the diverging lens. What is the focal length of the diverging lens?

For the converging lens,
f = 15 cm
u = 60 cm

1/v + 1/u = 1/f
1/v = 1/f - 1/u
1/v = (u - f)/(uf)
v = u f/(u - f) = 60 * 15/(60 - 15)
v = 20 cm

For the diverging lens,
u = 30 cm - 20 cm = 10 cm
v = -6 cm
1/f = 1/v + 1/u
f = u v/(u + v)
f = 10 * (-6)/(10-6)
f = -15 cm (Answer)

Note - I used the sign convention in which distance to real are taken positive, and distance to imaginary as negative. Let me know if you are using some other sign convention.
I used u for object distance and v for image distance. Replace if you use some other symbols.

I suggest you draw the diagram.

akhl IF-Rockerz
akhl
akhl

Joined: 30 March 2007
Posts: 5275

Posted: 05 March 2012 at 1:54am | IP Logged
Originally posted by Miggi

Q the angle of incidence of a ray of light in air is adjusted gradually as it enters a shallow tank made of plexigals and filled with carbon disulfide. The tank is open at the top -the light ray passes directly  from air into the carbon disulfide. Is there a minimum angle of incidence for which light is transmitted into the carbon disulfide but not into te plexigals at the bottom of the tank? If so find the angle. If not explain why not? Index of refraction values are air 1.00, plexigals 1.51, carbon disulfide 1.628. 


Let theta1 = angle with normal in air
theta2 = angle with normal in carbon disulfide

Critical angle between carbon disulfide and plexigals = sin-1(1.51/1.628) = 68.1 deg

sin(theta1)/sin(theta2) = 1.51
sin(theta1) = 1.51 * sin(theta2)
theta1 is minimum when theta2 is minimum. Minimum value of theta2 = 68.1 deg

sin(theta1) = 1.51 * sin(68.1 deg)
sin(theta1) = 1.4
This is not possible because the value of sin(theta1) must be between -1 and +1

Therefore, it is not possible to fulfull the given requirement.

.SilentPrincess IF-Rockerz
.SilentPrincess
.SilentPrincess

BollyCurry Screen Writer
Joined: 31 March 2009
Posts: 7953

Posted: 05 March 2012 at 8:06am | IP Logged
I'm doing this project... on how it is to be youg in norway today.

Were we're supposed to interview some people and collect all data and present it.

My question is.. what kind of questions should I ask? Like what questions should I ask to know as much as possible about that person?
Any help?

Would be glad if any suggestion would come before wednesday!

Thank you
.SilentPrincess

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