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3 years ago

Which technology is intended to work at a distance of about 5 to 10 centimeters with transmission speeds of 250 Kbps?

Physics

1 answer:

Mars2501 [29]3 years ago

5 0

Answer:

NFC Near Field Communication

Explanation:

The Near Field Communication is a communication protocol, for extra short distance with a maximum of 10 centimeters, but usually used in 4 to 5 cm. Its intended to be used in contactless pay systems and in transportation card. Actually has been used to transfer multimedia from cell phones and other devices. The maximum data rate is around 424 kbit/s, with mean in 250 Kbp

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What is the formula to find how far a wheel travels in one rotation?

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it is also known as formula for circumfrence which is 2 times pi times radius. if radius was 5 then the circumfrence would be 10pi.

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3 years ago

A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele

evablogger [386]

Answer:

a)n= 3.125 x $10^{19$ electrons.

b)J= 1.515 x $10^{6$ A/m²

c) $V_{d$ =1.114 x $10^{4$ m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x $10^{-3$ m

radius 'r' = d/2 => 1.025 x $10^{-3$ m

no. of electrons 'n'= 8.5 x $10^{28}$

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x $10^{-19$ C

n= Q/e => 5/ 1.6 x $10^{-19$

n= 3.125 x $10^{19$ electrons.

b) the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x $10^{-3$ )²)

J= 1.515 x $10^{6$ A/m²

c) The typical speed' $V_{d$ ' of an electron is given by:

$V_{d$ = $\frac{J}{n|q|}$

=1.515 x $10^{6$ / 8.5 x $10^{28}$ x |-1.6 x $10^{-19$ |

$V_{d$ =1.114 x $10^{4$ m/s

d) According to these equations,

J= I/A

$V_{d$ = $\frac{J}{n|q|}$ = $\frac{I}{nA|q|}$

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

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2 years ago

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Explain how Pascal's principle can be used to design a fluid power system and describe how a fluid power system works.

zhenek [66]

Pascal's law of fluid transfer states that when there is an increase in fluid pressure, the rest of the extrinsic variables also increases. For example, in a flow of liquid in an orifice, there is a contraction of diameter in the orifice part. The fluid that will go in there increases in pressure and thereby an increase in velocity as well.

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3 years ago

One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s

The mass of the identical balls are

$m = m_{1}=m_{2}$

(a). We need to calculate the speed of the first ball after the collision

Using law of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Along X- axis

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}$

$v_{1}=u_{1}+u_{2}$

Put the value into the formula

$v_{1}=2.2i-0.80i$

$v_{1}=1.4i\ m/s$

Along Y-axis

$0=m_{1}v_{1}+m_{2}v_{2}$

$m_{1}v_{1}=-m_{2}v_{2}$

$v_{1}=-v_{2}$

Put the value into the formula

$v_{1}=-1.36j\ m/s$

Then the final speed of the first ball

$v_{1}=\sqrt{(1.4)^2+(1.36)^2}$

$v_{1}=1.95\ m/s$

(b) We need to calculate the direction of the first ball after the collision

Using formula of direction

$\tan\theta=\dfrac{v_{2}}{v_{1}}$

$\tan\theta=\dfrac{-1.36}{1.4}$

$\theta=\tan^{-1}\dfrac{-1.36}{1.4}$

$\theta=-44.16^{\circ}$

Negative sign shows the direction of first ball .

Hence, (a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

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3 years ago

An airplane is flying through a thundercloud at a height of 2 000 m. (this is a very dangerous thing to do because of updrafts,

tester [92]

It is given that an<span> airplane is flying through a thundercloud at a height of 2000 m.
</span><span>
Since the parity of charges is opposite and the airplane lies between the two charges and both the electric fields are in the same direction at the plane. Therefore, the magnitudes of the electric field at the aircrafts will add up.
Now, check the image to see the calculations:

</span>

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3 years ago

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