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3 years ago

5

Describe how an electromagnet can be switched on and off.

1 answer:

Phoenix [80]3 years ago

8 0

Answer:The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.

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The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

3 years ago

From a set of graphed data the slope of the best fit line is found to be 1.35 m/s and the slope of the worst fit line is 1.29m/s

Svetradugi [14.3K]

Solution:

Let the slope of the best fit line be represented by ' $m_{best}$ '

and the slope of the worst fit line be represented by ' $m_{worst}$ '

Given that:

$m_{best}$ = 1.35 m/s

$m_{worst}$ = 1.29 m/s

Then the uncertainity in the slope of the line is given by the formula:

$\Delta m = \frac{m_{best}-m_{worst}}{2}$ (1)

Substituting values in eqn (1), we get

$\Delta m = \frac{1.35 - 1.29}{2}$ = 0.03 m/s

8 0

3 years ago

The first step in a star formation is ______?

Mariana [72]

First look at the star

8 0

3 years ago

Read 2 more answers

A circular loop of wire with radius 2.00 cm and resistance 0.600 Ω is in a region of a spatially uniform magnetic field B⃗ that

Anika [276]

Answer:

Incomplete questions

Let assume we are asked to find

Calculate the induced emf in the coil at any time, let say t=2

And induced current

Explanation:

Flux is given as

Φ=NAB

Where

N is number of turn, N=1

A=area=πr²

Since r=2cm=0.02

A=π(0.02)²=0.001257m²

B=magnetic field

B(t)=Bo•e−t/τ,

Where Bo=3T

τ=0.5s

B(t)=3e(−t/0.5)

B(t)=3exp(-2t)

Therefore

Φ=NAB

Φ=0.001257×3•exp(-2t)

Φ=0.00377exp(-2t)

Now,

Induce emf is given as

E= - dΦ/dt

E= - 0.00377×-2 exp(-2t)

E=0.00754exp(-2t)

At t=2

E=0.00754exp(-4)

E=0.000138V

E=0.138mV

b. Induce current

From ohms laws

V=iR

Given that R=0.6Ω

i=V/R

i=0.000138/0.6

i=0.00023A

i=0.23mA

5 0

3 years ago

What types of landforms can water form?

Alika [10]

Lakes and rivers can be formed landforms from water

4 0

3 years ago