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3 years ago

Joe follows the procedure below to make a compass.

Physics

2 answers:

Anna [14]3 years ago

8 0

The answer would be C.

koban [17]3 years ago

7 0

Answer:

C: to magnetize the needle

Explanation:

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A material you are testing conducts electricity but canot be pulled into wires

agasfer [191]

A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metalloid. Hope this helps!

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3 years ago

True or false question please help me out

Len [333]

Answer:

thats true

Explanation:

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2 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

3 years ago

Value of g is independent of

Margaret [11]

The lord of the greeks answer d

5 0

3 years ago

Read 2 more answers

In a certain time period a coil of wire is rotated from one orientation to another with respect to a uniform 0.38-T magnetic fie

juin [17]

<h2>The emf produced is 7.2 V</h2>

Explanation:

When coil is placed in the magnetic field , the flux attached with it can be found by the relation . Flux Ф = the dot product of magnetic field and area of coil .

Thus Ф = B A cosθ

here B is magnetic field strength and A is the area of coil .

The angle θ is the angle between coil and field direction .

When coil rotates , the angle varies . By which the flux varies . The emf is produced in coil due to variation of flux . The relation for this is

The emf produced ξ = - $\frac{d\phi}{dt}$ = B A sinθ $\frac{d\theta}{dt}$

Now in the given problem

5 = 0.38 x A x $\frac{d\theta}{dt}$ I

Now if the magnetic field is 0.55 T and all the other terms are same , the emf produced

ξ = 0.55 x A x $\frac{d\theta}{dt}$ Ii

dividing II by I , we have

$\frac{\xi}{5}$ = $\frac{0.55}{0.38}$ = 1.45

or ξ = 7.2 V

6 0

3 years ago