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1 year ago

The amount of straight pipe needed to bend a given radius. Also, the actual length of the conduit that will be bent.

1 answer:

8 0

Development length: The actual length of the bent conduit. Gain: Since a conduit bends radially rather than at an angle, the total length will not match the length required for all bends. Gain is the amount of space that is saved by a $90^{0}$ curve.

<h3>Bent Conduit</h3>

Conduit benders from Klein Tools are built to function and last longer than even the highest professional standards. To ensure a favourable experience and significantly enhance the final result of your project, it is advised that you become familiar with bending concepts, procedures, and the bender's capabilities. The benders are labelled with various alignment symbols to enable the operator make the bends required to complete any job. This aids bending while executing a ground or air bend. Arrow, teardrop, star point, and angle markings are the symbols on the Klein Tools benders. On certain bender head sides, you can see these markings.

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A relatively small impact crater 20 kilometers in diameter could be made by a comet 2 kilometers in diameter traveling at 25 km/

nadya68 [22]

Answer:

$KE=1.3\times 10^{21}\ J$

$KE=3.1\times 10^{5}\ E$

Explanation:

(a)

Given:

mass of comet, $m=4.2\times 10^{12}\ kg$

velocity of the comet, $v=2.5\times 10^4\ m.s^{-1}$

<u>Now, the kinetic energy of the comet can be given by:</u>

$KE=\frac{1}{2} m.v^2$

$KE=0.5\times 4.2\times 10^{12}\times (2.5\times 10^4)^2$

$KE=1.3\times 10^{21}\ J$

(b)

Given:

energy released by 1 megaton of TNT, $E=4.2\times 10^{15}\ J$

<u>Now the kinetic energy of the comet in terms of energy of 1 megaton TNT:</u>

$KE=\frac{1.3\times 10^{21}}{4.2\times 10^{15}} E$

i.e.

$KE=3.1\times 10^{5}\ E$

4 0

2 years ago

Un satélite geoestacionario se encuentra a una distancia de 120.000 km sobre la superficie de Júpiter. Determine: a. El periodo

Lisa [10]

Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

b) f = 2.7777 10⁻⁵ Hz

c) d) w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

T = 10 h (3600 s / 1h) = 3.6 104 s

b) the period the frequency are related

T = 1 / f

f = 1 / T

f = 1 / 3.6 104

f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

The distance traveled is equal to the length of the circumference

d = 2pi (R + r)

d = 2pi (69 911 103 + 120 106)

d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

.w = 2pi /t

w = 2pi / 3.6 10⁴

w = 1.745 10⁻⁴ rad / s

how fast is

v = w r

v = 1.75 10-4 (69.911 106 + 120 106)

v = 190017 m / s

5 0

3 years ago

List down all the jovian planets in order of increasing distance from the sun

Olenka [21]

Answer:

There are total eight planets in the solar system and the average distance from the sun to each planet in increasing order is given below.

Explanation:

The average distance from the sun is listed below in increasing order.

1. Mercury - It is the most closet planet to Sun, 57 million km

2. Venus - 108 million km

3. Earth - 150 million km

4. Mars - 228 million km

5. Jupiter - 779 million km

6. Saturn - 1.43 billion km

7. Uranus - 2.88 billion km

8. Neptun - It is the most farthest from the Sun, 4.50 billion km

8 0

3 years ago

If Phoenix, Arizona, experiences a cool, wet day in June, does that mean the regions climate is changing?

jenyasd209 [6]

It doesn't mean that the climate is changing, it is probably just the morning dew.

5 0

3 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago