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3 years ago

All waves travel through a

Physics

1 answer:

Brilliant_brown [7]3 years ago

5 0

Conductor answer is conductor I’m sure

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What is the best definition of

Brut [27]

Answer:

Work is best defined as a force exerted on a body to cause it move over a certain distance.

work=force×displacement×cosθ

Explanation:

7 0

2 years ago

20. I am a noble gas with 2 electrons.

madam [21]

Answer:

20.helium

21.silicon

hope this helps :)

8 0

4 years ago

What is the y component of a vector defined as 12.2m at 81.5°?

sergejj [24]

Answer:

Explanation:

This is a displacement vector since it is defined in terms of distance (meters, to be exact). The way you find the y-component is

$V_y=Vsin\theta$ which says that you multiply the magnitude of the vector (its length) by the sin of the direction (the angle):

$V_y=12.2sin(81.5)$ and get

$V_y=$ 12.1 m

3 0

3 years ago

If a circular loop of wire of radius 19.7 cm is located in a region where the spatially uniform magnetic field perpendicular to

Alisiya [41]

Answer:

Magnitude of induced emf will be $2.437\times 10^{-3}volt$

Explanation:

We have given radius of the circular loop r = 19.7 cm = 0.197 m

So area of the circular loop $A=\pi r^2=3.14\times 0.197^2=0.1218m^2$

It is given that magnetic field is changing as $2\times 10^{-3}T/sec$

Emf induced in the circular loop is equal to $e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}$

So emf induced will be equal to $e=\frac{-d\Phi }{dt}=-0.1218\times2\times 10^{-3}=2.437\times 10^{-3}volt$

So magnitude of induced emf will be $2.437\times 10^{-3}volt$

6 0

3 years ago

The wires in a household lamp cord are typically 3.5 mm apart center to center and carry equal currents in opposite directions.

zhuklara [117]

Explanation:

It is given that,

Distance between wires, d = 3.5 mm = 0.0035 m

Power of light bulb, P = 100 W

Potential difference, V = 120 V

(a) We need to find the force per unit length each wire of the cord exert on the other. It is given by :

$\dfrac{F}{l}=\dfrac{\mu_o I^2}{2\pi r}$

Power, P = V × I

$I=\dfrac{P}{V}=\dfrac{100}{120}=0.83\ A$

This gives, $\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times (0.83)^2}{2\pi \times 0.0035}$

$\dfrac{F}{l}=0.0000393\ N/m$

$\dfrac{F}{l}=3.93\times 10^{-5}\ N/m$

(b) Since, the two wires carry equal currents in opposite directions. So, teh force is repulsive.

(c) This force is negligible.

Hence, this is the required solution.

8 0

3 years ago