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miss Akunina [59]

2 years ago

15

What is the wavelength of an AM radio wave in a vacuum if its frequency is 810 kHz?

1 answer:

dalvyx [7]2 years ago

6 0

Answer- 370.114 meters

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When stopping at a railroad crossing, you should be no closer than?

emmainna [20.7K]

You should stop<span> before </span>crossing railroad<span> tracks: Whenever a </span>crossing<span> is not ... Follow </span>no closer than<span> 10 feet behind the large truck. </span>

7 0

3 years ago

The mean free path of a helium atom in helium gas at standard temperature and pressure is 0.2 um.What is the radius of the heliu

ivolga24 [154]

Answer: $0.10233nm$

Explanation:

The mean free path $\lambda$ of an atom is given by the following formula:

$\lambda=\frac{RT}{\sqrt{2} \pi d^{2}N_{A}P}$ (1)

Where:

$\lambda=0.2\mu m=0.2(10)^{-6}m$

$R=8.3145J/mol.K$ is the Universal gas constant

$T=0\°C=273.115K$ is the absolute standard temperature

$d$ is the diameter of the helium atom

$N_{A}=6.0221(10)^{23}/mol$ is the Avogadro's number

$P=1atm=101.3kPa=101.3(10)^{3}Pa=101.3(10)^{3}J/m^{3}$ absolute standard pressure

Knowing this, let's find $d$ from (1), in order to find the radius $r$ of the helium atom:

$d=\sqrt{\frac{RT}{\sqrt{2}\pi\lambda N_{A}P}}$ (2)

$d=\sqrt{\frac{(8.3145J/mol.K)(273.115K)}{\sqrt{2}\pi(0.2(10)^{-6}m)(6.0221(10)^{23}/mol)(101.3(10)^{3}J/m^{3})}}$ (3)

$d=2.0467(10)^{-10}m$ (4)

If the radius is half the diameter:

$r=\frac{d}{2}$ (5)

Then:

$r=\frac{2.0467(10)^{-10}m}{2}$ (6)

$r=1.0233(10)^{-10}m$ (7)

However, we were asked to find this radius in nanometers. Knowing $1nm=(10)^{-9}m$ :

$r=1.0233(10)^{-10}m.\frac{1nm}{(10)^{-9}m}=0.10233nm$ (8)

Finally:

$r=0.10233nm$ This is the radius of the helium atom in nanometers.

5 0

2 years ago

A square coil (length of side = 24 cm) of wire consisting of two turns is placed in a uniform magnetic field that makes an angle

dalvyx [7]

Explanation:

It is given that,

Length of side of a square, l = 24 cm = 0.24 m

The uniform magnetic field makes an angle of 60° with the plane of the coil.

The magnetic field increases by 6.0 mT every 10 ms. We need to find the magnitude of the emf induced in the coil. The induced emf is given by :

$\epsilon=N\dfrac{d\phi}{dt}$

$\dfrac{d\phi}{dt}$ is the rate of change if magnetic flux.

$\phi=BA\ cos\theta$

$\theta$ is the angle between the magnetic field and the normal to area vector.

$\theta=90-60=30$

$\epsilon=NA\dfrac{dB}{dt}\times cos30$

$\epsilon=2\times (0.24\ m)^2\times \dfrac{6\ mT}{10\ mT}\times cos(30)$

$\epsilon=0.0598\ T$

$\epsilon=59.8\ mT$

or

EMF = 60 mT

So, the magnitude of emf induced in the coil is 60 mT. Hence, this is the required solution.

6 0

3 years ago

What is abiotic in science and what is biotic in science

zhuklara [117]

Abiotic is anything that is not alive (i.e. Rock)
Biotic is anything that is alive (I.e. Sunflower)

3 0

2 years ago

A parallel plate capacitor with capacitance C is connected to a power supply AV, acquiring a charge on its plates. While it is s

Sonja [21]

Answer:

b) ${{Q}'}=2.1Q$

Explanation:

Given that

The power source still connects it means that the voltage difference will be same in above both condition

As we know that

$Q=C\Delta V$

Or we can say that

$Q=\dfrac{\varepsilon _oA}{d}\Delta V$ ----1

Where d is the distance between two plates ,A is the area.

When K=4.2 and distance become double

${Q}'=\dfrac{K\varepsilon _oA}{{d}'}\Delta V$

${Q}'=\dfrac{4.2\varepsilon _oA}{2d}\Delta V$ ----2

Now from equation 1 and 2

$\dfrac{{Q}'}{Q}=\dfrac{4.2}{2}$

So

${{Q}'}=2.1Q$

8 0

3 years ago