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ivanzaharov [21]

3 years ago

12

Which device is used to increase or decrease the voltage of an alternating

1 answer:

Kazeer [188]3 years ago

6 0

Answer:

D.

A transformer

Explanation:

Transformer is used to regulate voltage

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Si el número masico del elemento litio es 7 ¿cual es su nucleo?

sp2606 [1]

El núcleo tiene 3 protones (lo que le da al núcleo una carga de +3, identificándolo como el elemento Litio) y 4 neutrones (lo que le da un número total de masa de 7).

4 0

3 years ago

The particles that are found in the nucleus of an atom are

kvv77 [185]

Answer: Electrons are the smallest of the three particles that make up atoms. Electrons are found in shells or orbitals that surround the nucleus of an atom

Explanation: hope this helps

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3 years ago

Air bubbles being released when breathing from the mouth

Inessa [10]

Answer:

air embolism

Explanation:

8 0

2 years ago

Need help finding the average speed.

Snowcat [4.5K]

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.38
1.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/3
0.95 + 1.61 + 0.56 = 3.12 / 3 = 1.04
0.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26

5 0

4 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

3 years ago