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Consider this situation: A rope is used to lift an actor upward off the stage. Of the forces listed, identify which act upon the

Katarina [22]

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6 0

4 years ago

Read 2 more answers

Change of a liquid to a solid

Hoochie [10]

Melting and frezzing

3 0

3 years ago

1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com

Vsevolod [243]

Answer and explanation:

A.

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

B.

In muon rest frame it travels Zero meters

C.

Distance, d = Velocity, v * Time, s

$where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s$

$d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m$

D.

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

E.

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

$d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}$

$d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}$

$d' = 594\sqrt{1 - 0.81}$

$d' = 594 \times 0.4359$

d' = 258.92m

F.

in the rest frame of someone standing on the mountain,

muon moves straight down

3 0

4 years ago

What is the frequency for a beam of electrons orbiting in a field of 4.62 x 10^-3 T? Let the mass of an electrons m = 9.31 x 10^

melisa1 [442]

Frequency: $1.27\cdot 10^8 Hz$

Explanation:

The force experienced by an electron in a magnetic field is

$F=qvB$

where

$q=1.6\cdot 10^{-19}C$ is the electron charge

v is the speed of the electron

B is the strength of the magnetic field

Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:

$qvB=m\frac{v^2}{r}$

where

r is the radius of the orbit

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

Re-arranging the equation,

$\frac{v}{r}=\frac{qB}{m}$ (1)

We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):

$v=\frac{2\pi r}{T}$

Substituting into (1),

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

We also know that 1/T is equal to the frequency f, so

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

In this problem,

$B=4.62\cdot 10^{-3}T$

Therefore, the frequency of the electrons is

$f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz$

4 0

3 years ago

An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the

Wewaii [24]

Answer:

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

Explanation:

As we know that the power emitted by the source is given as

$P = 170 W$

now we know that

$P = \frac{N}{t} (\frac{hc}{\lambda})$

now we know that energy density is given as

$u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}$

now we have

$E = B c$

$u = \frac{B^2}{2\mu_0}$

intensity is defined as

$I = \frac{P}{A}$

now we have

$\frac{I}{c} = u = \frac{B^2}{2\mu_0}$ [/tex]

now we have

$\frac{dB}{dt} = \omega B$

$\frac{dB}{dt} = \frac{2\pi c B}{\lambda}$

$\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}$

here we have

$I = \frac{P}{4\pi r^2}$

$I = \frac{170}{4\pi (410)^2}$

$I = 8.05 \times 10^{-5}$

now we have

$\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}$

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

4 0

3 years ago