Ask question

Ask question

All categories

3 years ago

10

1. How far away must you be from a 675 kHz radio station with power 50.0 kW for there to be only one photon per second per squar

e meter? Assume no reflections or absorption, as if you were in deep outer space. 2. Discuss the implications for detecting intelligent life in other solar systems by detecting their radio broadcasts.

1 answer:

goldenfox [79]3 years ago

6 0

a) 0.321 ly

b) 0.321 light years is not far in astronomical terms. Alien life would need to transmit at tremendous power in order for their radio transmissions to be detectable. Their radio signal also needs to be stronger than background noise in order to be distinguishable. Therefore it is unlikely that radio transmissions from alien life will ever be detected.

You might be interested in

Atoms are happy (they will not readily react with other elements) when they have a full outside ring of

Len [333]

Answer: TRUE

Explanation:

Atoms are happy when they will not react with other elements while having a full outside ring of electrons because this makes them to be noble.

A stable atom possesses full outside ring of electrons while unstable one does not. So, they are happy also because of stability.

7 0

3 years ago

Which type of boundary is modeled?

Katena32 [7]

A is convergent
B is

7 0

3 years ago

A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The

marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0

3 years ago

Physics B 2020 Unit 3 Test

weqwewe [10]

Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

$F=qvB sin \theta$

where here:

For the proton in this problem:

$q=1.602\cdot 10^{-19}C$ is the charge of the proton

v = 300 m/s is the speed of the proton

B = 19 T is the magnetic field

$\theta=65^{\circ}$ is the angle between the directions of v and B

So the force is

$F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N$

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the force experienced by a charged particle moving in a magnetic field.

It can be applied as follows:

- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

In this problem:

- Direction of motion = to the right (index finger)

- Direction of field = downward (middle finger)

- Direction of force = into the screen (thumb)

4)

The radius of a particle moving in a magnetic field is given by:

$r=\frac{mv}{qB}$

where here we have:

$m=6.64\cdot 10^{-22} kg$ is the mass of the alpha particle

$v=2155 m/s$ is the speed of the alpha particle

$q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C$ is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

$r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m$

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

$f=\frac{qB}{2\pi m}$

where here:

$q=1.602\cdot 10^{-19}C$ is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

$m=9.31\cdot 10^{-31} kg$ is the mass of the electron

Substituting, we find:

$f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz$

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

Therefore in this case, when the particle gains the opposite charge, the circular motion 2) changes sign, so the path will remains helical, but it reverses direction.

7)

The electromotive force induced in a conducting loop due to electromagnetic induction is given by Faraday-Newmann-Lenz:

$\epsilon=-\frac{N\Delta \Phi}{\Delta t}$

where

N is the number of turns in the loop

$\Delta \Phi$ is the change in magnetic flux through the loop

$\Delta t$ is the time elapsed

From the formula, we see that the emf is induced in the loop (and so, a current is also induced) only if $\Delta \Phi \neq 0$ , which means only if there is a change in magnetic flux through the loop: this occurs if the magnetic field is changing, or if the area of the loop is changing, or if the angle between the loop and the field is changing.

8)

The flux is calculated as

$\Phi = BA sin \theta$

where

B = 5.5 T is the strength of the magnetic field

A is the area of the coil

$\theta=18^{\circ}$ is the angle between the direction of the field and the plane of the loop

Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

$A=(0.15 m)(0.08 m)=0.012 m^2$

So the flux is

$\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb$

See the last 7 answers in the attached document.

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> docx </span>

<span class="sg-text sg-text--link sg-text--bold sg-text--link-disabled sg-text--blue-dark"> pdf </span>

5 0

3 years ago

An interstellar space probe is launched from Earth. After a brief period of acceleration it moves with a constant velocity, 70.0

sleet_krkn [62]

Answer:

22.26 years

, 15.585 light years , 11.13 light years

Explanation:

a)

$t' = t/(\sqrt{1-(v/(c*v)/c)}$

= $15.9/\sqrt{(1-0.7*0.7)}$

= 22.26 years

b)

0.7*c*22.26 years

=15.585 light years

c)

0.7*c*15.9

=11.13 light years

3 0

3 years ago