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pashok25 [27]
2 years ago
7

"The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radiu

s 3 centimeters at the bottom of a cylindrical tank of radius 20 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the accelertion of gravity, which you should approximate as 1000 cm/s2
The first equation says that the gain in kinetic energy of the water leaving the tank equals the loss in potential energy of the water in the tank.
1/2 delta m v2 = delta m gh
The second equation says that the rate at which water leaves the tank equals the rate of decrease in the volume of water in the tank (which is conservation of mass because water has constant density)
pi 102^2 dh/dt = pi 3^2 v
Derive a differential equation for the height of water in the tank.
If the initial height of the water is 30 centimeters, find a formula or the solution.
According to the model, how long does it take to empty the tank?
Another way to solve this differential equation is to make the substitution w = underroot h. What is the differential equation that w satisfies?
Physics
1 answer:
Leno4ka [110]2 years ago
5 0

Answer:

The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radius 3 centimeters at the bottom of a cylindrical tank of radius 10 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the acceleration of gravity, which you should approximate as 1000 cm/s2.

shdh

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Suppose that a barometer was made using oil with rho=900 kg/m3. What is the height of the barometer at atmospheric pressure?
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Hey there!

The pressure under a liquid column can be , calculated using  the following formula :

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Read 2 more answers
A moving sidewalk 95 m in length carries passengers at a speed of 0.53 m/s. One passenger has a normal walking speed of 1.24 m/s
Archy [21]

Answer:

a) t = 1.8 x 10² s

b) t = 54 s

c) t = 49 s

Explanation:

a) The equation for the position of an object moving in a straight line at constan speed is:

x = x0 + v * t

where

x = position at time t

x0 = initial position

v = velocity

t = time

In this case, the origin of our reference system is at the begining of the sidewalk.

a) To calculate the time the passenger travels on the sidewalk without wlaking, we can use the equation for the position, using as speed the speed of the sidewalk:

x = x0 + v * t

95 m = 0m + 0. 53 m/s * t

t = 95 m/ 0.53 m/s

t = 1.8 x 10² s

b) Now, the speed of the passenger will be her walking speed plus the speed of th sidewalk (0.53 m/s + 1.24 m/s = 1.77 m/s)

t = 95 m/ 1.77 m/s = 54 s

c) In this case, the passenger is located 95 m from the begining of the sidewalk, then, x0 = 95 m and the final position will be x = 0. She walks in an opposite direction to the movement of the sidewalk, towards the origin of the system of reference ( the begining of the sidewalk). Then, her speed will be negative ( v = 0.53 m/s - 2*(1.24 m/s) = -1.95 m/s. Then:

0 m = 95 m -1.95 m/s * t

t = -95 m / -1.95 m/s = 49 s

3 0
2 years ago
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