Which type of electromagnetic radiation is useful in communications technology?

A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

marta [7]

This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

The current I is any motion of charge from one region to another, so this is given by:

 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

The magnitude of the current density is:

$J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})$

Being:

$A=2mm^{2} = 2x10^{-6}m^{2}$

Finally, for the drift velocity magnitude vd, we find:

 $v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)$

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd

6 0

2 years ago

Un objeto se suelta desde determinada altura y emplea un tiempo t en caer al suelo. Si se cuadruplica la altura desde la cual se

blondinia [14]

When an object falls from a h height, you should work with the uniformly accelerated linear movement equations:

y=½*a*t²+Vo*t+yo

You should consider:

a=-g=-10m/s²

yo=h

If it’s a freefall, it means it starts from rest, which means it has no initial velocity:

Vo=0

Replacing that information in the equation:

y=½*(-10m/s²)*t²+0*t+h=-5m/s²*t²+0+h=-5m/s²*t²+h

So this is the

Besides, if you want to find out how long it takes for it to get to the floor, you should put the height of the floor as final height, which would be 0 (assuming the initial height has been measured from there):

y=0

0=-5m/s²*t²+h

5m/s²*t²=h

t²=h/(5m/s²)

t=√(h/(5m/s²))

t=√(hs²/(5m))

t=(√(h/(5m)))s

If we quadruple h:

t2=(√(h2/(5m)))s=(√(4*h1/(5m)))s=(√4)*(√h1/(5m)))s=2*(√h1/(5m)))s=2*t1

This 4 goes inside the square root, so then it converts to 2. So the new time is twice as much the previous time.

Concerning velocity, you have to use the other equation:

v=at+vo

As I said before, a is gravity and vo is zero.

v=-10m/s²*t+0=-10m/s²*t

Final velocity is directly related to time, so if time is doubled, so is velocity.

v2=-10m/s²*t2=-10m/s²*(2*t1)=2*(-10m/s²*t1)=2*v1

So the correct answer is A, and the other ones are false.

8 0

3 years ago

URGENT PLEASE HELP!!!!

AlexFokin [52]

It just completely stops kinetic energy is movement so technically it would stop.

6 0

3 years ago

a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric

n200080 [17]

Part a)

Here in this we can use momentum conservation as there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}$

here we know that

$m_1 = 0.004 kg$

$v_{1i} = 650 m/s$

$m_2 = 0.095$

$v_{2i} = 0$

$v_{2f} = 23 m/s$

now by above equation

$0.004*650 + 0.095* 0 = 0.004*v + 0.095*23$

$2.6 + 0 = 0.004*v + 2.185$

$v = 103.75 m/s$

Part b)

Final kinetic energy of the system

$KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

$KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2$

$KE_f = 46.65 J$

Initial Kinetic energy of the system will be

$KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2$

$KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2$

$KE_f = 845 J$

So here kinetic energy is decreased for this system

final energy is less than initial energy

6 0

2 years ago

A friend says that the reason one's hair stands out while touching a charged Van de Graaff generator is simply that the hair str

Llana [10]

Answer:

Explanation:

Yes I agree with the statement .

When a person who is perfectly insulated from the earth , touches a Van de Graaff , his body acquires charge . when the hair acquires it, it stands out due to mutual repulsion . It is to be noted here that at pointed areas on a surface , there is larger accumulation of charge. Accumulation of charge is greater at hair tops .

It is also a general observation that when a bird sits on high tension wire , his feather stands out due to the same reason.

5 0

2 years ago