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2 years ago

13

Help ASAP I’ll mark you as brainlister

1 answer:

Sonja [21]2 years ago

8 0

Answer:

The Moon
Mars
The ISS ( The international space station)

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An antigen is a protein made by your body to respond to a specific foreign molecule.

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hope this helps it's F

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3 years ago

two pulse waves a and b arrive at a point in a medium simustaneously. if the amplitude of wave a is 35 units the amplitude of wa

Shalnov [3]

It depends on the type of interference.
For constructive interference, add the amplitudes to get |35 + 41| = 76 units.
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2 years ago

A ball is dropped from rest at the top of a 6.10 m

natita [175]

Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

$(\frac{1}{\sqrt{10} })^n=\frac{2.38}{6.1}$ = .00396

Taking log on both sides

- n / 2 = log .00396

n / 2 = 2.4

n = 5 approx

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2 years ago

Which statement accurately describes the outer planets?

tiny-mole [99]

The outer planets have a high gravity due to their large size

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3 years ago

Read 2 more answers

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

2 years ago