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3 years ago

10

A jetliner is traveling east from Salt Lake City to Washington DC, a distance of 1,850 miles. The jetliner travels at an average

air speed of 590 mph relative to the air. At the jetliner's altitude, the wind is blowing 36 mph to the west. How long does the journey take?

1 answer:

maks197457 [2]3 years ago

3 0

The jetliner is traveling against the wind. The net speed of the jetliner is
590 mph - 36 mph = 554 mph
The time it takes for the jetliner to arrive at the destination is
1850 miles / 554 mph = 3.34 hours

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Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. Part A If the surfac

alukav5142 [94]

Answer:

5308.34 N/C

Explanation:

Given:

Surface density of each plate (σ) = 47.0 nC/m² = $47\times 10^{-9}\ C/m^2$

Separation between the plates (d) = 2.20 cm

We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:

$E=\dfrac{\sigma}{2\epsilon_0}$

Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,

$E_{between}=E+E=2E=\frac{2\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Now, plug in $47\times 10^{-9}\ C/m^2$ for 'σ' and $8.85\times 10^{-12}\ F/m$ for $\epsilon_0$ and solve for the electric field. This gives,

$E_{between}=\frac{47\times 10^{-9}\ C/m^2}{8.854\times 10^{-12}\ F/m}\\\\E_{between}= 5308.34\ N/C$

Therefore, the electric field between the plates has a magnitude of 5308.34 N/C

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2 years ago

1hp is written in electric motor what does it mean

Angelina_Jolie [31]

Explanation:

The horsepower (hp) is a unit in the foot-pound-second ( fps ) or English system, sometimes used to express the rate at which mechanical energy is expended. It was originally defined as 550 foot-pounds per second (ft-lb/s). A power level of 1 hp is approximately equivalent to 746 watt s (W) or 0.746 kilowatt s (kW).

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3 years ago

One string of a certain musical instrument is 70.0 cm long and has a mass of 8.79 g . It is being played in a room where the spe

Svetach [21]

To solve this problem we will apply the concepts of linear mass density, and the expression of the wavelength with which we can find the frequency of the string. With these values it will be possible to find the voltage value. Later we will apply concepts related to harmonic waves in order to find the fundamental frequency.

The linear mass density is given as,

$\mu = \frac{m}{l}$

$\mu = \frac{8.79*10^{-3}}{70*10^{-2}}$

$\mu = 0.01255kg/m$

The expression for the wavelength of the standing wave for the second overtone is

$\lambda = \frac{2}{3} l$

Replacing we have

$\lambda = \frac{2}{3} (70*10^{-2})$

$\lambda = 0.466m$

The frequency of the sound wave is

$f_s = \frac{v}{\lambda_s}$

$f_s = \frac{344}{0.768}$

$f_s = 448Hz$

Now the velocity of the wave would be

$v = f_s \lambda$

$v = (448)(0.466)$

$v = 208.768m/s$

The expression that relates the velocity of the wave, tension on the string and linear mass density is

$v = \sqrt{\frac{T}{\mu}}$

$v^2 = \frac{T}{\mu}$

$T= \mu v^2$

$T = (0.01255kg/m)(208.768m/s)^2$

$T = 547N$

The tension in the string is 547N

PART B) The relation between the fundamental frequency and the $n^{th}$ harmonic frequency is

$f_n = nf_1$

Overtone is the resonant frequency above the fundamental frequency. The second overtone is the second resonant frequency after the fundamental frequency. Therefore

$n=3$

Then,

$f_3 = 3f_1$

Rearranging to find the fundamental frequency

$f_1 = \frac{f_3}{3}$

$f_1 = \frac{448Hz}{3}$

$f_1 = 149.9Hz$

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3 years ago

The figure above shows 4forces 3N, 10N, 3√3N, and 6N acting on a particle P. The resultant of the four forces is.

Brilliant_brown [7]

As you gave no pic I took them on one lined

F_1=3N
F_2=10N
F_3=3root 3 N
F_4=6N

$\\ \sf\longmapsto F_{net}=F_1+F_2\dots$

$\\ \sf\longmapsto F_{net}=3+10+6+3\sqrt{3}$

$\\ \sf\longmapsto F_{net}=19+3(1.732)$

$\\ \sf\longmapsto F_{net}=19+5.196$

$\\ \sf\longmapsto F_{net}=24.196N$

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2 years ago

Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV

forsale [732]

Answer:

(a) $\lambda=1.227\ A$

(b) $\lambda=0.388\ A$

(c) $\lambda=0.038\ A$

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$

Where

m and e are the mass of and charge on an electron

On solving,

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

V = 100 V

$\lambda=\dfrac{12.27}{\sqrt{100} }\ A$

$\lambda=1.227\ A$

(b) V = 1 kV = 1000 V

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

$\lambda=\dfrac{12.27}{\sqrt{1000} }\ A$

$\lambda=0.388\ A$

(c) If $V=100\ kV=10^5\ V$

$\lambda=\dfrac{12.27}{\sqrt{10^5} }\ A$

$\lambda=0.038\ A$

Hence, this is the required solution.

7 0

2 years ago