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3 years ago

What do microwaves have in common with light waves?

Physics

2 answers:

frutty [35]3 years ago

7 0

The answer is C because viable light and microwaves are both on the electromagnetic spectrum glad to helpelelec

Rom4ik [11]3 years ago

4 0

Hey you!

I believe the correct answer would be C. Both are electromagnetic waves.

I Really Do Hope This Is Correct, And That It Helps You!

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A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic

defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0

3 years ago

G a magnetic field perpendicular to the plane of a wire loop is uniform in space but changes with time t in the region of the lo

Luden [163]

Answer:

e = Δφ / Δt induced emf is proportional to enclosed flux

Also φ = B * A flux is proportional to area and enclosed field

If the induced emf e increases with time than the flux and hence the magnetic field is increasing with time (replace B with G)

Since e = ΔG * A / Δt if e is linear then G must also be linear and be proportional to the time

6 0

1 year ago

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Dvinal [7]

Mexico Mexico language to is the answer to the question

3 0

3 years ago

Which component of a galaxy is often found between the stars and looks like a cloud or smoke?

slava [35]

Answer:

Hey I would say Dust

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5 0

2 years ago

Brayden and Riku now use their skills to work a problem. Find the equivalent resistance, the current supplied by the battery and

Liono4ka [1.6K]

a) $5 \Omega$ , 1.6 A

b) $6 \Omega$ , 1.33 A

Explanation:

In this situation, we have two resistors connected in series.

The equivalent resistance of resistors in series is equal to the sum of the individual resistances, so in this circuit:

$R=R_1+R_2$

where

$R_1=4\Omega$

$R_2=1 \Omega$

Therefore, the equivalent resistance is

$R=4+1=5 \Omega$

Now we can use Ohm's Law to find the current flowing through the circuit:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=5\Omega$ is the equivalent resistance of the circuit

Substituting,

$I=\frac{8}{5}=1.6 A$

The two resistors are connected in series, therefore the current flowing through each resistor is the same, 1.6 A.

In this part, a third resistor is added in series to the circuit; so the new equivalent resistance of the circuit is

$R=R_1+R_2+R_3$

where:

$R_1=4\Omega\\R_2=1\Omega\\R_3=1\Omega$

Substituting, we find the equivalent resistance:

$R=4+1+1=6 \Omega$

Now we can find the current through the circuit by using again Ohm's Law:

$I=\frac{V}{R}$

where

V = 8 V is the voltage supplied by the battery

$R=6\Omega$ is the equivalent resistance

Substituting,

$I=\frac{8}{6}=1.33 A$

And the three resistors are connected in series, therefore the current flowing through each resistor is the same, 1.33 A.

3 0

3 years ago