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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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If 2 objects are moved by the same force (F):
kipiarov [429]

Answer:y=mx+b 58+5

Explanation:

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2 years ago
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If 20 beats are produced within a single second, which of the following frequencies could possibly be held by two sound waves tr
zhuklara [117]

The correct choice is

D. 22 Hz and 42 Hz.

In fact, the beat frequency is given by the difference between the frequencies of the two waves:

f_B = |f_1 -f_2|

In this problem, the beat frequency is f_B=20 Hz, therefore the only pair of frequencies that gives a difference equal to 20 Hz is

D. 22 Hz and 42 Hz.

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3 years ago
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Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400
hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(c) The beetle reaches a height of 2.1 m.

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

v^2=u^2+2as

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s

The launching speed of the beetle is <u>6.4 m/s</u>.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

v=u+at

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms

The time taken by the beetle to launch itself upwards is <u>1.62 ms</u>.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m

The beetle can jump to a height of <u>2.1 m</u>



7 0
3 years ago
I need it now aaaasssssaaaappp!!!!
Vera_Pavlovna [14]

A) Pitch depends on the frequency of a sound wave. The higher the pitch, the higher the frequency and the lower the pitch, the lower the frequency. Sound waves have different pitches due to varying frequency levels.

B) Sound is created as the pen vibrates. This vibrations also interacts with the air atoms and molecules, causing them to vibrate too, therefore, creating sound waves. The reason the brain grows annoyed at this continual sound is because the brain will focus on this sound only, causing the brain to go into overdrive, creating annoyance.

C) Sound waves are travelling vibrations of particles. Space does not have any atoms or particles for the sound vibrations to interact with, therefore creating zero sound waves to travel.

Hope this helps!! :))

5 0
2 years ago
You are packing for a trip to another star. During the journey, you will be traveling at 0.99c. You are trying to decide whether
Elenna [48]

Answer:

Do neither of these things ( c )

Explanation:

For length contraction : Is calculated considering the observer moving at a speed that is relative the object at rest applying this formula

L = (l) \sqrt{1 -\frac{v^{2} }{c^{2} } }

where l = Measured distance from object at rest, L =  contracted measured in relation to the observer , v = speed of clock , c = speed of light

you will do neither of these things because before you can make such decisions who have to view the object in this case yourself from a different frame from where you are currently are, if not your length and width will not change hence you can't make such conclusions/decisions .

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