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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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A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t
arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vectorn=-0.60\hat{i}-0.080\hat{j}

We need to calculate the torque on the loop

Using formula of torque

\tau=NIAB\sin\theta

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}

\tau=4.7\times10^{-4}\ N-m

Hence, The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

5 0
2 years ago
The frequency of the applied RF signal used to excite spins is directly proportional to the magnitude of the static magnetic fie
Anni [7]

Answer:

The inverse frequency is \dfrac{3}{80}\ s

Explanation:

Given that,

Magnetic field = 20 T

Proportionality constant = 5 Hz/T

Change in magnetic field = 3 T

We know that,

B=\dfrac{K}{\dfrac{1}{\omega}}

We need to calculate the inverse frequency

Using formula of frequency

\Delta(\dfrac{1}{\omega})=\dfrac{\Delta B}{k\times(\dfrac{1}{\omega^2})}

\Delta(\dfrac{1}{\omega})=\dfrac{k\times\Delta B}{B^2}

Put the value into the formula

\Delta(\dfrac{1}{\omega})=\dfrac{3\times5}{(20)^2}

\Delta(\dfrac{1}{\omega})=\dfrac{3}{80}\ s

Hence, The inverse frequency is \dfrac{3}{80}\ s

5 0
2 years ago
A car travels three-quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. the initial vel
schepotkina [342]
20.3 divided by 3.0 will get u velocity and v times 3.0s 
3 0
2 years ago
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
Airida [17]

Answer:

126.99115 g

Explanation:

50 g at 90 cm

Stick balances at 61.3 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

The mass of the meter stick is 126.99115 g

6 0
2 years ago
Read 2 more answers
As the drawing illustrates, a siren can be made by blowing a jet of air through 20 equally spaced holes in a rotating disk. The
Aneli [31]

Answer:

ω = 630.2663 = 630[rad/s]

Explanation:

Solution:

- We can tackle this question by simple direct proportion relation between angular speed for the disk to rotate a cycle that constitutes 20 holes. We will use direct relation with number of holes per cycle to compute the revolution per seconds i.e frequency of speed f.

                                  1rev(20 hole) -> 20(cycle)/rev  

                                        2006.2(cycle) -> f ?  

                              f = 2006.2/20 = 100.31rev at second  

- The relation between angular frequency and angular speed is given by:

                                 ω = 2πf

                                 ω = 2*3.14*100.31

                                 ω = 630.2663 = 630[rad/s]

4 0
2 years ago
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