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nataly862011 [7]

2 years ago

14

A continuous function y = ƒ(x) is known to be negative at x = 0 and positive at x = 1. Why does the equation ƒ(x) = 0 have at le

ast one solution between x = 0 and x = 1? Illustrate with a sketch

1 answer:

exis [7]2 years ago

3 0

Answer:

69

Explanation:

69420=420659782

fjkdod

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Find the current Lx in the figure

AleksandrR [38]

Explanation:

$\frac{1}{8} + \frac{1}{2} \\ 1.6 + 1.4 = 3 \\ \frac{1}{3} + \frac{1}{9} \\ 2.25 + 2 = 4.25 \: ohm$

R total = 4.25 ohm

I total = Vt/Rt

I total= 17/4.25= 4 A

Ix= 600 mA

$\frac{9}{9 + 3} \times 4 = 3\\ \frac{2}{2 + 8} \times 3 = 0.6a \\ = 0.6 \: milli \: amper$

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3 years ago

The best grade of hardwood lumber that is generally available is _____

Vesnalui [34]

Answer:

FAS

Explanation:

first and second grade

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2 years ago

Find the velocity and acceleration of box B when point A moves vertically 1 m/s and it is 5 m

Goshia [24]

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7 0

3 years ago

Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water

olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate = 2 L/s

water temperature 30 degree celcius

$Q = A\times V$

$2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity$

$V = \frac{20\times 4}{9\times \pi} = 2.83 m/s$

$Re = \frac{\rho\times V\times D}{\mu}$

at 30 degree celcius $= \mu = 0.798\times 10^{-3}Pa-s , K = 0.6154$

$Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}$

Re = 106390

So ,this is turbulent flow

$Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}$

$Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419$

$\frac{h\times 0.03}{0.615} = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}$

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 0

3 years ago

Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a

Arada [10]

Answer:

a) $h_c = 0.1599 W/m^2-K$

b) $H_{loss} = 5.02 W$

c) $T_s = 302 K$

d) $\dot{Q} = 25.125 W$

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, $T_a = 25^0 C = 298 K$

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

$\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600} = 1.25 kg/sec$

Rate of heat transfer,

$\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W$

a) To calculate the convection coefficient relationship for heat transfer by convection:

$\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K$

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, $k_c = 0.8$

$\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K$

b) Heat loss from the pipe to the environment:

$H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W$

d) The required fan control power is 25.125 W as calculated earlier above

5 0

3 years ago