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3 years ago

6

An ocean wave has an amplitude of 2.5 m. Weather conditions suddenly change such that the wave has an

1 answer:

Dovator [93]3 years ago

8 0

Answer:

2.5

Explanation:2.5 +2.5 = 5.0

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This spectrometer reading shows some red, blue, and purple. Our atom is most likely ______________________.

Kisachek [45]

<span>This spectrometer reading shows some red, blue, and purple. Our atom is most likely Hydrogen source.

This spectrometer reading shows some reds, orange, and yellow. Our atom is most likely Neon source.

This spectrometer reading shows some red, yellow, and blue. Our atom is most likely Helium source.

This spectrometer reading shows some yellow, blue, and purple. Our atom is most likely Mercury source</span>

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2 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

An ideal heat engine absorbs 97.2 kJ of heat and exhausts 83.8 kJ of heat in each cycle. What is the efficiency of the engine?

r-ruslan [8.4K]

Answer:

13.7%

Explanation:

Given that,

Heat absorbed by the engine = 97.2 kJ

Heat exhausted by the engine in each cycle = 83.8 kJ

We need to find the efficiency of the engine. It is calculated by the formula.

$\eta=1-\dfrac{Q_e}{Q_a}\\\\=1-\dfrac{83.8}{97.2}\\\\=0.137\\\\=13.7\%$

so, the efficiency of heat engine is 13.7%.

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3 years ago

You push a box with a force of 80 n. if the net force on the box is 50 n, what is the force on the box due to sliding friction?

Elodia [21]

The force of the sliding friction is 30 N.

3 0

3 years ago

A machine part has the shape of a solid uniform sphere of mass 205 g and diameter 4.10 cm. It is spinning about a frictionless a

kondor19780726 [428]

Answer:

Answer is in the following attachment.

Explanation:

3 0

3 years ago