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2 years ago

What’s the answer to this I don’t understand

Engineering

1 answer:

kari74 [83]2 years ago

3 0

Answer:

Electrons are the negatively charged particles of atom. Electrons are extremely small compared to all of the other parts of the atom. So, the negative green particles of this atom are the electrons.

Explanation:

Hope it helps:)

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1. A turbine in a steam power plant operates isentropically with an inlet pressure (P3) of 3.5 MPa and inlet temperature (T3) of

sergey [27]

Answer:

See attached picture.

Explanation:

See attached picture for explanation.

4 0

3 years ago

From the following numbered list of characteristics, decide which pertain to (a) precipitation hardening, and which are displaye

Blababa [14]

Answer:

(a) Precipitation hardening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.

(2) The hardening/strengthening effect is not retained at elevated temperatures for this process.

(4) The strength is developed by a heat treatment.

(b) Dispersion strengthening

(1) The strengthening mechanism involves the hindering of dislocation motion by precipitates/particles.

(3) The hardening/strengthening effect is retained at elevated temperatures for this process.

(5) The strength is developed without a heat treatment.

7 0

3 years ago

A satellite is launched 600 km from the surface of the earth, with an initial velocity of 8333.3 m./s, acting parallel to the ta

Vikki [24]

Answer:

eccentrcity of orbit is 0.22

Explanation:

GIVEN DATA:

Initial velocity of satellite = 8333.3 m/s

distance from the sun is 600 km

radius of earth is 6378 km

as satellite is acting parallel to the earth therefore $\theta angle = 0$

and radial component of given velocity is zero

we have $h = r_o v_r_o = 6378+600 =6.97*10^6 m$

h = 6.97*10^6 *8333.3 = 58.08*10^9 m^2/s

we know that

$\frac{1}{r} =\frac{GM}{h^2} \times ( i + \epsilon cos\theta)$

$GM = gr^2 = 9.81*(6.37*10^6)^2 = 398*10^{12} m^3/s$

$\frac{1}{6.97*10^6} =\frac{398*10^{12}}{(58.08*10^9)^2} \times ( i + \epsilon cos0)$

solvingt for $\epsilon)$

$\epsilon = 0.22)$

therefore eccentrcity of orbit is 0.22

6 0

3 years ago

In the fully developed region of flow in a circular pipe, does the velocity profile change in the flow direction?

taurus [48]

Answer:

<em>No, the velocity profile does not change in the flow direction.</em>

Explanation:

In a fluid flow in a circular pipe, the boundary layer thickness increases in the direction of flow, until it reaches the center of the pipe, and fill the whole pipe. If the density, and other properties of the fluid does not change either by heating or cooling of the pipe, <em>then the velocity profile downstream becomes fully developed, and constant, and does not change in the direction of flow.</em>

3 0

3 years ago

For a cylindrical annulus whose inner and outer surfaces are maintained at 30 ºC and 40 ºC, respectively, a heat flux sensor mea

miskamm [114]

Answer:

$k=0.12\ln(r_2/r_1)$ $\frac {W}{ m^{\circ} C}$

where $r_1$ and $r_2$ be the inner radius, outer radius of the annalus.

Explanation:

Let $r_1$ , $r_2$ and $L$ be the inner radius, outer radius and length of the given annulus.

Temperatures at the inner surface, $T_1=30^{\circ}C\\$ and at the outer surface, $T_2=40^{\circ}C$ .

Let q be the rate of heat transfer at the steady-state.

Given that, the heat flux at r=3cm=0.03m is

$40 W/m^2$ .

$\Rightarrow \frac{q}{(2\pi\times0.03\times L)}=40$

$\Rightarrow q=2.4\pi L \;W$

This heat transfer is same for any radial position in the annalus.

Here, heat transfer is taking placfenly in radial direction, so this is case of one dimentional conduction, hence Fourier's law of conduction is applicable.

Now, according to Fourier's law:

$q=-kA\frac{dT}{dr}\;\cdots(i)$