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alexgriva [62]
3 years ago
10

Water flows at 0.850 m/s from a hot water heater, through a 450-kPa pressure regulator. The pressure in the pipe supplying an up

stairs bathtub 3.70m above the heater is 414-kPa. What's the flow speed in this pipe?
Physics
1 answer:
grin007 [14]3 years ago
4 0

Answer:

The velocity is  v_2= 0.45 \ m/s

Explanation:

From the question we are told that

      The initial speed of the hot water is  v_1 = 0.85 \ m/s

     The pressure from the heater  P_1  =  450 \ KPa = 450 *10^{3} \ Pa

      The height of the hot water before flowing is  h_1 = 0 \ m

      The height of bathtub above the heater is h_2  =  3.70 \ m

       The pressure in the pipe is P_2 =  414 KPa = 414 *10^{3} \ Pa

       The density of water is \rho =  1000 \ kg/m^3

Apply Bernoulli equation

      P_1 + \rho gh_1 +\frac{1}{2} \rho v_1^2  =  \rho g h_2 + \frac{1}{2}\rho v_2 ^2

Substituting values

     (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2)  =  (1000 * *9.8*3.70) + (0.5*1000*v_2^2 )

=>   v_2^2 =  \frac{ (450 *10^{3}) + (1000 * 9.8 *0 ) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}

=>   v_2= \sqrt{ \frac{ (450 *10^{3}) + (1000 * 9.8 * 0) + (0.5 * 1000 * 0.85^2) -[ (1000 * *9.8*3.70) ]}{0.5*1000}}

=>    v_2= 0.45 \ m/s

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Answer:

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Explanation:

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Compare the time period of two simple pendulums of length 4m and 16m at a place.
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Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

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Read 2 more answers
Three joggers are running along straight lines as follows: Jogger A, with a mass of 55.2kg, is traveling along the line y=6.00m
frutty [35]

Answer:

L = - 1361.591 k Kgm/s

Explanation:

Given

mA = 55.2 Kg

vA = 3.45 m/s

rA = 6.00 m

mB = 62.4 Kg

vB = 4.23 m/s

rB = 3.00 m

mC = 72.1 Kg

vC = 4.75 m/s

rC = - 5.00 m

then we apply the equation

L =  (mv x r)

⇒  LA = mA*vA x rA = 55.2 *(3.45 i)x(6 j) = (1142.64 k) Kgm/s

⇒  LB = mB*vB x rB = 62.4 *(4.23 j)x(3 i) = (- 791.856 k) Kgm/s

⇒  LC = mC*vC x rC = 72.1 *(- 4.75 j)x(- 5 i) = (- 1712.375 k) Kgm/s

Finally, the total counterclockwise angular momentum of the three joggers about the origin is

L = LA + LB + LC = (1142.64 - 791.856 -1712.375) k  Kgm/s

L = - 1361.591 k Kgm/s

7 0
3 years ago
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