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4 years ago

Which travels at the greatest speed?

1 answer:

5 0

"Light through a vacuum" is the one among the following choices given in the question that shows which can travel at the greatest speed. The correct option among all the options that are given in the question is the fourth option or option "D". I hope that this is the answer that has come to your desired help.

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f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If

Luden [163]

Answer:

The angular velocity is $w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

The Radial acceleration which is mathematically represented as

$a_r = \frac{v^2}{r} = w^2r$

And the Tangential acceleration which is mathematically represented as

$a_r = \alpha r$

The net acceleration is evaluated as

$a = \sqrt{a_r^2 + a_t^2}$

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

So at maximum angular acceleration we a have

$a_{max} = \sqrt{a_r^2 + a_t^2}$

$a_{max}^2 = a_r^2 + a_t^2$

$a_{max}^2 = (w^2r)^2 + (\alpha r)^2$

$a_{max}^2 = r^2 w^4 + r^2 \alpha ^2$

$a_{max}^2 = r^2 (w^4 + \alpha^2 )$

$w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}$

$w^4 = \frac{a_{max}^2}{r^2} - \alpha ^2$

$w= \sqrt[4]{\frac{a_{max}^2}{r^2} - \alpha ^2}$

3 0

3 years ago

Why are the magnetic fields of superconducting magnets often stronger than those of conventional magnets?

g100num [7]

The superconducting magnets are able to generate powerful magnetic fields because they have no electrical resistance.

To find the answer, we have to know more about the superconducting magnets.

<h3>What is superconducting magnet?</h3>

An example of an electromagnet is a superconducting magnet.
They are constructed from coils of superconducting wire and must be used while being chilled to cryogenic temperatures.
Because the wire encircling the magnet has no electrical resistance when it is in its superconducting condition, they may produce powerful magnetic fields.
Because of this, the magnet can conduct far greater electrical currents than the typical electromagnet.

Thus, we can conclude that, the superconducting magnets are able to generate powerful magnetic fields because they have no electrical resistance.

Learn more about the superconducting magnets here:

brainly.com/question/1476682

#SPJ4

3 0

2 years ago

Two resistors with values of 6.0 W and 12 W are connected in parallel. This combination is connected in series with a 2.0 W resi

Murljashka [212]

Well, first of all, you really shouldn't use ' W ' for the unit when you
talk about resistors.

You may have seen the resistors written as 6ω, 12ω, and 2ω in your
book or on the homework sheet. But that little symbol ' ω ' is not a ' w '.
It's the small Greek letter 'omega'. The CAPITAL omega is ' Ω '. It's used
to label resistors because it's short for "ohms". So the resistors in this
problem have resistances of 6Ω, 12Ω, and 2Ω, and we have to do some
manipulating of the individual resistors to find out what resistance the
battery actually sees.

The parallel combination of the first two resistors looks like a single
resistor, whose value is

1 / (1/6 + 1/12)

= 1 / (2/12 + 1/12)

= 1 / (3/12)

= 12/3 = 4Ω .

Now, that parallel combination is connected in series with 2Ω .
All three resistors together look like a single resistor of

4Ω + 2Ω = 6Ω .

So the battery thinks there's a single resistor connected to it,
with 6Ω of resistance. The current out of the battery is

I = V / R = (24v) / (6Ω) = 4 Amperes.

That 4 Amperes of current will split between the parallel resistors,
but it will ALL flow through the series 2Ω resistor because there's
no other path through that part of the circuit.

So the current through the 2Ω resistor is 4 Amperes. (B).

Note:
The POWER dissipated by the 2Ω resistor is

P = I² R = (4A)² · (2Ω) = 32 watts .

This is a fair amount of heat, so you'll need to provide some way
to remove the heat from the resistor, otherwise it'll burn or crack.

8 0

3 years ago

A 15-ft3 tank contains oxygen initially at 14.7 psia and 80°F. A paddle wheel within the tank is rotated until the pressure insi

Nikolay [14]

Explanation:

Equation for energy balance will be as follows.

$\Delta E_{system} = E_{in} - E_{out}$

$\Delta U = W_{in} - Q_{out}$

Hence, $W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Therefore, we will calculate the final temperature as follows.

$\frac{P_{1}V}{T_{1}} = \frac{P_{2}V}{T_{2}}$

$T_{2} = \frac{20 psia}{14.7}(638 R)$

= 868.03 R

Now, we will calculate the mass as follows.

m = $\frac{P_{1}V}{RT_{1}}$

= $\frac{14.7 psia \times 15 ft^{3}}{0.3353 psi ft^{3}/lbm R \times 638 R}$

= 1.031 lbm

Hence,

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

Putting the values into the above equation as follows.

$W_{in} = Q_{out} + mC_{v} (T_{2} - T_{1})$

$W_{in} = 20 Btu + 1.031 lbm (\frac{0.160 Btu}{lbm R})(735 - 540)R$

$W_{in}$ = 655.2 Btu

Thus, we can conclude that work done by paddle wheel is 655.2 Btu.

6 0

3 years ago

. A long 10-cm-diameter steam pipe whose external surface temperature is 110oC passes through some open area that is not protect

Nata [24]

Answer:

$Nu = 30.311$

Explanation:

Let consider that pipe is a horizontal cylinder. The Nusselt number is equal to:

$Nu = \left\{0.6+\frac{0.387\cdot Ra_{D}^{\frac{1}{6} }}{[1+\left(\frac{0.559}{Pr} \right)^{\frac{9}{16}} ]^{\frac{8}{27} }} \right\}^{2}$ , for $Ra_{D} \le 10^{12}$ .

Where $Ra_{D}$ is the Rayleigh number associated with the cylinder.

The Rayleigh number is:

$Ra_{D} = \frac{g\cdot \beta\cdot (T_{pipe}-T_{air})\cdot D^{3}}{\nu^{2}}\cdot Pr$

By assuming that air behaves ideally, the coefficient of volume expansion is:

$\beta = \frac{1}{T}$

$\beta = \frac{1}{283.15\,K}$

$\beta = 3.532\times 10^{-3}\,\frac{1}{K}$

The cinematic and dynamic viscosities, thermal conductivity and isobaric specific heat of air at 10 °C and 1 atm are:

$\nu = 1.426\times 10^{-5}\,\frac{m^{2}}{s}$

$\mu = 1.778\times 10^{-5}\,\frac{kg}{m\cdot s}$

$k = 0.02439\,\frac{W}{m\cdot ^{\textdegree}C}$

$c_{p} = 1006\,\frac{J}{kg\cdot ^{\textdegree}C}$

The Prandtl number is:

$Pr = \frac{\mu\cdot c_{p}}{k}$

$Pr = \frac{(1.778\times 10^{-5}\,\frac{kg}{m\cdot s} )\cdot (1006\,\frac{J}{kg\cdot ^{\textdegree}C} )}{0.02439\,\frac{W}{m\cdot ^{\textdegree}C} }$

$Pr = 0.733$

Likewise, the Rayleigh number is:

$Ra_{D} = \frac{(9.807\,\frac{m}{s^{2}} )\cdot (3.532\times 10^{-3}\,\frac{1}{K} )\cdot (110^{\textdegree}C-10^{\textdegree}C)\cdot (0.1\,m)^{3}}{(1.426\times 10^{-5}\,\frac{m^{2}}{s})^{2} }\cdot (0.733)$

$Ra_{D} = 12.486\times 10^{6}$

Finally, the Nusselt number is:

$Nu = \left\{0.6+\frac{0.387\cdot (12.486\times 10^{6})^{\frac{1}{6} }}{\left[1 + \left(\frac{0.559}{0.733}\right)^{\frac{9}{16} }\right]^{\frac{8}{27} }} \right\}^{2}$

$Nu = 30.311$

8 0

3 years ago

Read 2 more answers