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aleksley [76]
2 years ago
11

Spherical water tank of radius R = 5m is emptied through a small circular hole of radius r = 0.03 m at the bottom. The top of th

e tank is open to the atmosphere. The instantaneous water level h in the tank (measured from the bottom of the tank, at the drain) can be determined from the solution of the following ODE:
dh /dt =r²(2gh)^0.5/ 2hR-h²
where g = 9.81 m/s². If the initial (t = 0) water level is h=6.5 m, compute the time required to drain the tank to a level of h= 0.5m. Use the fourth-order Runge-Kutta method.
Mathematics
1 answer:
tatuchka [14]2 years ago
7 0

Answer:

water level is h=6.5 m, compute the time required to drain the tank to a level of h= 0.5m. Use the fourth-order Runge-Kutta method.

Step-by-step explanation:

water level is h=6.5 m, compute the time required to drain the tank to a level of h= 0.5m. Use the fourth-order Runge-Kutta method.

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2 years ago
A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 65°F room. After fift
s344n2d4d5 [400]

86.34 minutes was used   to cool the soup to 80 °F

<h3></h3><h3>How can we calculate this using Newton's Law of Cooling?</h3>

Newton's Law of Cooling can be computed as :

T=T1+(T0- T1) e^{kt}

The initial temperature  is been represent by T0 which is  ( 100 F)

The temperature of surrounding  is been given as  (69 F)

The  final temperature  is been given  as  T

t = time

we were told that the temperature of the soup move to  95 F, using  15 minutes.

Hence,  T = 95 F

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Then we can input the values into the above equation as :

95= 69 + (100-69)e^{15k}

if we simplify this , we have

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Then e^{15k} =0.8387

Then we can find the value of k as :

k= -0.012

Then we have the new equation as

T= 69 + (100-69)e^{-0.012t}

T= 69 + 31e^{15k}

Then at T=80F

we have,

80= 69 + 31e^{15k}

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1 year ago
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3 years ago
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3 years ago
Need help solving this !
Zigmanuir [339]

Answer:

Step-by-step explanation:

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