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Fiesta28 [93]
1 year ago
12

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant

urinates at a remarkable rate of 0.0060 m3 (a bit over a gallon and a half) per second. Assume that the urine exits 1.0 m below the bladder and passes through the urethra, which we can model as a tube of diameter 8.0 cm and length 1.2 m. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.
1) What is the speed of the flow?
2) If we assume that the liquid is at rest in the bladder (a reasonable assumption) and that the pressure where the urine exits is equal to atmospheric pressure, what does Bernoulli's equation give for the pressure in the bladder?
Physics
1 answer:
weqwewe [10]1 year ago
8 0

Answer:

a) v =  1.19 m / s , b)   P₁ = 0.922 10⁵ Pa

Explanation:

1) Let's use the fluid continuity equation

       Q = A v

The area of ​​a circle is

      A = π r2 = π d²/4

     

     v = Q / A = Q 4 / pi d²

     v = 0.006 4/π 0.08²

     v =  1.19 m / s

2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point

     P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂

The exercise tell us

P₂ = 1.0013 105 Pa

v₁ = 0

y₁ = 1 m

y₂=0  

Rho (water) = 1000 kg / m³

      P₁ + rho y₁ = P₂ + ½ rho v₂²

      P₁ = P₂ + ½ rho v₂² - rho g y₁

      P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 10⁵ +708.5  - 9800

      P₁ =  92208.5Pa

      P₁ = 0.922 10⁵ Pa

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Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency  of pipe is given by

f=\frac{nv}{2L}

where L=length of Pipe

v=velocity of sound

f=150\ Hz for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

f=\frac{(2n+1)}{4L}\cdot v

for n=0

f_1=\frac{v}{4L}

f_1=150\ Hz

for n=2

f_2=\frac{5v}{4L}

Thus there is one additional resonance corresponding to n=1 , between f_1 and f_2

8 0
2 years ago
A worker wants to turn over a uniform 1110-N rectangular crate by pulling at 53.0 ∘ on one of its vertical sides (the figure (Fi
tekilochka [14]
This problem has three questions I believe:

> How hard does the floor push on the crate?

<span>We have to find the net vertical (normal) Fn force which results from Fp and Fg. 
We know that the normal component of Fg is just Fg, which is equal to as 1110N. 
From the geometry, the normal component of Fp can be calculated: 
Fpn = Fp * cos(θp) 
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The total normal force Fn then is: 
Fn = Fg + Fpn 
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= 1721.63 N</span>

 

> Find the friction force on the crate

<span>We have to look for the net horizontal force Fh which results from Fp and Fg. Since Fg is a normal force entirely,  so we can say that the horizontal component is zero: 

Fh = Fph + Fgh 
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= 1016.31 N * sin(53) 
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> What is the minimum coefficient of static friction needed to prevent the crate from slipping on the floor?

We just need to compute the ratio Fh to Fn to get the minimum μs.

 

μs = Fh / Fn

= 811.66 N / 1721.63 N

<span>= 0.47</span>

8 0
2 years ago
The A-string (440 HzHz) on a piano is 38.9 cmcm long and is clamped tightly at both ends. If the string tension is 667N, what's
Mice21 [21]

Answer:

Mass, m = 2.2 kg                                

Explanation:

It is given that,

Frequency of the piano, f = 440 Hz

Length of the piano, L = 38.9 cm = 0.389 m

Tension in the spring, T = 667 N

The frequency in the spring is given by :

f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}

\mu=\dfrac{m}{L} is the linear mass density

On rearranging, we get the value of m as follows :

m=\dfrac{T}{4Lf^2}

m=\dfrac{667}{4\times 0.389\times (440)^2}

m = 0.0022 kg

or

m = 2.2 grams

So, the mass of the object is 2.2 grams. Hence, this is the required solution.

3 0
2 years ago
Read 2 more answers
The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You
tatiyna

Answer:

Explanation:

Given

B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)

Em wave is in the form of

B=B_0\sin (kx-\omega t)

where \omega =frequency\ of\ oscillation

k=wave\ constant

B_0=Maximum\ value\ of\ Magnetic\ Field

Wave constant for EM wave k is

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Wavelength of wave \lambda =\frac{2\pi }{k}

\lambda =\frac{2\pi }{1.05\times 10^7}

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7 0
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Suppose that sunlight is incident upon both a pair of reading glasses and a pair of sunglasses. Which pair would you expect to b
Ainat [17]

Answer: the pair of sunglasses

Explanation:

A good pair of sunglasses are composed of abosorbent lenses that filter the sunlight that affects the eyes retina, especially ultraviolet (UV). So, these sunglasses are used to reduce the amount of light or radiant energy transmitted.

On the other hand, normal reading glasses (in which the lens glass has not been treated to filter ultraviolet sunlight) will let UV rays pass through.

Therefore, if both glasses are exposed to sunlight, the sunglasses are expected to be warmer by absorbing that radiant energy and preventing it from reaching the eyes.

4 0
1 year ago
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