How do you keep the weight evenly distributed on a trailered boat?

A radio antenna 326 m in diameter receives a radio signal from a very distant object at perpendicular incidence. The radio signa

Fed [463]

Assume that the antenna absorbs all the radiation that falls on the disk and calculate the power in Watts received by the antenna are the (a) 3.0e-8.

<h3>What is perpendicular with examples?</h3>

Two awesome traces intersecting every different at 90° or a proper perspective are referred to as perpendicular traces. Here, AB is perpendicular to XY due to the fact AB and XY intersect every different at 90°. The traces are parallel and do now no longer intersect every different. They can in no way be perpendicular to every different.

Read more about the perpendicular :

brainly.com/question/1202004

#SPJ1

6 0

2 years ago

Carbon dioxide contained in a piston–cylinder device is compressed from 0.3 to 0.1 m3. During the process, the pressure and volu

Lunna [17]

Answer:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$

Explanation:

For this case we know the following info :

$V_i = 0.3 m^3$ represent the initial volume

$V_f = 0.1 m^3$ represent the final volume

We know that the pressure and volume are related with the following expression:

$P = aV^{-2}= \frac{a}{V^2}$

Where a is a constant given $a = 6.5 Kpa m^6$

And we need to calculate the work associated to this process.

We have a compression, and by definition the work is defined with the following general expression:

$W = \int_{V_i}^{V_f} P dV$

If we replace the expression for P we got:

$W = \int_{V_i}^{V_f} a V^{-2} dV$

If we integrate we got:

$W = -a (\frac{1}{V}) \Big|_{V_i}^{V_f}$

Using the fundamental theorem of calculus we have:

$W = -a (\frac{1}{V_f} -\frac{1}{V_i})$

And replacing the values we got:

$W = -6.5 Kpa m^6 (\frac{1}{0.1 m^3}- \frac{1}{0.3 m^3})= -6.5 Kpa m^6 (6.66 m^-3) =-43.33 Kpa m^3 = -43.33 KJ$

5 0

4 years ago

A 240 V, 60 Hz squirrel-cage induction motor has a full-load slip of 0.02 and a full-load speed of 1764 rpm. The winding resista

earnstyle [38]

Answer:

full load current = 7.891151 A

so correct option is b. Ir= 7.89

Explanation:

given data

Energy E = V = 240 V

frequency = 60 Hz

full-load slip = 0.02

full-load speed = 1764 rpm

winding resistance = 0.6 Ω

winding reactance = 5 Ω

to find out

rotor current at full-load

solution

we will apply here full load current formula that is express as

full load current = $\frac{S*E}{\sqrt{R^2+ (S*X)^2}}$ ...................1

here S is full-load slip and E is energy given and R is winding resistance and X is winding reactance

put here value we get

full load current = $\frac{0.02*240}{\sqrt{0.6^2+ (0.02*5)^2}}$

full load current = 7.891151 A

so correct option is b. Ir= 7.89

6 0

3 years ago

A rectangular beam having b=300 mm and d=575 mm, spans 5.5 m face to face of simple supports. It is reinforced for flexure with

katovenus [111]

Answer:

provide 180 mm spacing

Explanation:

GIVEN DATA

rectangular beam: (b) = 300 mm, (d) = 575 mm

reinforced for flexure = 4Ф32 bars

WD = 30 kN /m, WL = 45 kN/m

Wu = 1.4 * 30 + 1.6 * 45 = 114 kN/m

i) concrete shear stress ( vc)

100 Ac / bd = (100 * u * $\frac{\pi }{4}$ * 32^2) / 300 * 575 = 1.865

from table 3.8

when: 100 Ac / bd = 1.865 then Vc = 0.778 N/mm^2

Ultimate shear force = (114 *5.5) / 2 = 313.5 kN

design shear stress = V / bd = (313.5 * 10^3) / (300 * 575) = 1.82 N/mm^2

v < 0.8 $\sqrt{22}$ = 1.82 < 3.75

design link provided according to

Asv / sv = b(v-vc) / 0.87 fy = 300(1.82 - 0.778) / 0.87 (420)

ASv / Sv = 0.855

From table 3.13 :the value of Asv / sv can be calculated as

$\frac{0.855 - 0.785}{0.897 - 0.785} = \frac{x - 200}{175 - 200}$

x = (-25) [ 0.625] + 200 = 184.375 mm

provide 180 mm spacing

8 0

4 years ago

Select the correct answer. Jude is a mechanical engineer. He works in the automobile industry. He is creating a prototype of an

Nikolay [14]

Answer:

Bid bureau of investigation

6 0

3 years ago