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3 years ago

12

Write in powers of 10 notation: 3 trillion; five-thousandths; 730,000,000,000,000; 0.000000000082.

1 answer:

Leno4ka [110]3 years ago

5 0

3 trillion= 3X10^<span>12
5 thousanths= 5X10^-3
730000000000000= 7.3X10^14
0.000000000082= 8.2X10^-11</span>

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A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

3 years ago

When sunlight strikes the side of a building, what form of energy is it<br> transformed to?

algol [13]

Answer:

thermal energy

Explanation:

8 0

2 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago

A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and

victus00 [196]

Answer:

a) $t=195.948N.m$

b) $\phi=13.6 \textdegree$

Explanation:

From the question we are told that:

Density $\rho=1.225kg/m^2$

Velocity of wind $v=14m/s$

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient $\mu=2.05$

a)

Generally the equation for Force is mathematically given by

$F=\frac{1}{2}\muA\rhov^2$

$F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2$

$F=163.29$

Therefore Torque

$t=F*r*sin\theta$

$t=163.29*1.2*sin90$

$t=195.948N.m$

b)

Generally the equation for torque due to weight is mathematically given by

$t=d*Mg*sin90$

Where

$d=sin \phi$

Therefore

$t=sin \phi*Mg*sin90$

$195.948=833sin \phi$

$\phi=sin^{-1}\frac{195.948}{833}$

$\phi=13.6 \textdegree$

5 0

3 years ago

If a bus travels 6 km in 10 minutes, what distance does it travel in 1 second

loris [4]

Answer:

Your answer is 10 m/s...

4 0

2 years ago

Read 2 more answers