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4 years ago

11

A garden hose with a small puncture is stretched horizontally along the ground. The hose is attached to an open water faucet at

one end and a closed nozzle at the other end. As you might suspect, water pressure builds up and water squirts vertically out of the puncture to a height of 0.93 m. Determine the pressure inside the hose. (Enter your answer to the nearest 1000 P

1 answer:

Marat540 [252]4 years ago

3 0

Answer:110.439 KPa

Explanation:

Given

Height to which water is raised $h=0.93 m$

assuming $P_{atm}=101.325 KPa$

Applying Bernoulli's equation between Pipe Puncture and highest Point we get

$P_1+0.5\rho v_1^2+\rho gh_1=P_2+0.5\rho v_2^2+\rho gh_2$

$v_1=0,h_1=0,v_2=0$

we get

$P_1-P_{atm}=\rho gh$

$\Delta P=10^3\times 9.8\times 0.93$

$P_1=P_{atm}+9.114$

$P_1=101.325+9.114=110.439 KPa$

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In a downhill ski race, surprisingly, little advantage is gained by getting a running start. (This is because the initial kineti

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Part a)

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Part b)

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Part a)

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Part b)

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then we will have

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 2.5^2 = 2(4.6)(69)$

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3 years ago

2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after

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Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

$A(t)=A_oe^{\frac{-bt}{2m}}$

or amplitude for unforce oscillation

$\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}$

or

$\frac{6bt}{2m}=1$

or

$b=\frac{m}{3T}$

Now, provided in the question Amplitude of the driven oscillation

$A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}$

the value of the maximum amplitude is obtained $(k=m\omega_d^2)$

thus,

$A=\frac{F_{max}}{(b\omega_d}$

Now, for n people on the bridge

Fmax = nF

thus,

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$0.075=\frac{nF}{((\frac{m}{3T})2\pi}$

or

n = 1810

hence, there were 1810 people on the bridge

b) $A=\frac{F_{max}}{(b\omega_d}$

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thus,

b=3b

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$A=\frac{F_{max}}{(3b\omega_d}$

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$A=\frac{1}{(3}A_o$

or

$A=\frac{1}{(3}0.075$

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A = 0.025 m = 25 mm

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