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guapka [62]
3 years ago
9

Please help me with the correct answer​

Physics
1 answer:
Rudik [331]3 years ago
3 0

Answer: fluid fraction

Explanation:

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PLS HELP WILL MARK BRAINLIEST
MariettaO [177]

Answer:

2.835 Watts

Explanation:

P = I²R

P = 1.5² × 1.26

P = 2.835 Watts

3 0
3 years ago
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List the ocean floor features that are formed by the movement of tectonic plates
Andreas93 [3]
Tsunami and under water volcano
8 0
3 years ago
What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?
vfiekz [6]

Answer: 0.0039\ A

Explanation:

Given

Diameter of the rod d=2.54\ cm

length of rod is l=20\ cm

Resistivity of silicon is \rho=6.4\times 10^2\ \Omega-m

cross-section of the rod A

\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2

Resistance of rod is  R

\Rightarrow R=\dfrac{\rho l}{A}

\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega

Current is given by

\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A

3 0
2 years ago
Una placa rectangular de metal tiene una longitud 8.430 m. y una anchura de 5.201 m. calcular el área de la placa con el número
bezimeni [28]

Responder:

43.84 m²

Explicación:

Dada la dimensión de una placa rectangular como:

Longitud (L) = 8.430 m

Ancho (W) = 5.201 m

El Área de una placa rectangular se puede obtener usando la fórmula:

Área = Largo × Ancho

Área = 8.430m * 5.201m

Superficie = 43.84443 m²

Área = 43.84 m²

3 0
3 years ago
Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is c_p=1.0\ Btu/lbm.F, and the steam properties as, A-4E:

h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
3 years ago
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