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3 years ago

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What is characteristic of both sound waves and electromagnetic waves?

1 answer:

Cerrena [4.2K]3 years ago

6 0

Sound waves are mechanical waves while electromagnetic waves are not. Thus, sound waves require a medium for propagation while electromagnetic waves do not. Sound waves are produced by mechanical vibrations and propagate through liquid, air and plasma as longitudinal waves. While electromagnetic waves are <span>produced by </span>accelerating or decelerating<span> charged particles. It also contain a</span> <span>magnetic field. Because they are not mechanical waves, it can propagate through a vacuum as well as air, liquid, and solid.</span>

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If 6000 J of heat is added to 200 gm of water at 25° C. What will be its final<br>temperature?

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Answer:

T₂ = 305.17 K

Explanation:

Given that,

Heat, Q = 6000 J

Mass, m = 200 gram

Initial temperature, T₁ = 25° C

We need to find its final temperature. Let it is T₂.

We know that,

$Q=mc\Delta T$

Where

c is the specific heat of water, c = 4.18 J/g°C

So,

$6000=200\times 4.18\times (T_2-298)\\\\\dfrac{6000}{200\times 4.18}=(T_2-298)\\\\7.17=(T_2-298)\\\\7.17+298=T_2\\\\T_2=305.17\ K$

So, the final temperature is equal to 305.17 K.

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I) A circular coil with radius 20 cm is placed with it's plane parallel and between two straight

Mamont248 [21]

Answer:

$B=15.433\ T$ inwards when viewing from the left side.

Explanation:

Given:

radius of the circular loop, $r=0.2\ m$

current in the coil, $I_c=0.5\ A$

Direction of current is clockwise when viewed from left.

Distance of wire P from the loop, $d_p=0.4\ m$

Distance of wire Q from the loop, $d_q=0.8\ m$

current in each wires P & Q, $I=0.2\ A$

Now the magnetic field in coil will be inwards when viewed from left by the Maxwell's right hand thumb rule.

<u>Magnitude is given by:</u>

$B_c=\frac{\mu_0.I_c}{2R}$

$B_c=\frac{4\pi\times 10^{-7}\times (0.5)}{2\times 0.2}$

$B_c=15.7\times 10^{-7}\ T$

<u>Now the effect of magnetic field due to wire P at the center of the loop:</u>

(We get the effective distance as 0.4+0.2=0.6 m)

$B_P=\frac{\mu_0.I}{2\pi.d_p}$

$B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 0.6}$

$B_P=6.67\times 10^{-8}\ T$ coming out of the loop when viewed from left.

<u>Now the effect of magnetic field due to wire Q at the center of the loop:</u>

(We get the effective distance as 0.8+0.2=1 m)

$B_P=\frac{\mu_0.I}{2\pi.d_q}$

$B_P=\frac{4\pi\times 10^{-7}\times (0.2)}{2\pi\times 1}$

$B_P=4\times 10^{-8}\ T$ going in to the loop when viewed from left.

<u>Now the net resultant effect all the magnetic fields:</u>

$B=B_c-B_P+B_Q$

$B=15.7-0.667+0.4$

$B=15.433\ T$ inwards when viewing from the left side.

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3 years ago