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zhuklara [117]
3 years ago
12

A transverse wave has an amplitude of 2.4m. What is the vertical distance, in meters, between the top of a Crest and the bottom

of a Trough?​
Physics
1 answer:
Alex17521 [72]3 years ago
3 0

Answer:

4.8 m

Explanation:

A transverse wave is a wave in which the oscillation of the particles occur in a direction perpendicular to the direction of motion of the wave.

The amplitude of a transverse wave is the maximum displacement of the wave with respect to the equilibrium position.

The crest of a wave is the position of maximum displacement on the positive side, while the trough of a wave is the position of maximum displacement on the negative side.

This means that the amplitude of a wave is equal to the distance between a crest (or a trough) and the equilibrium position.

Therefore, we have:

d=2A

where:

d is the distance between the top of a crest and the bottom of a trough

A is the amplitude of the wave

Here we have:

A = 2.4 m

So, we find

d=2\cdot 2.4 = 4.8 m

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Two scales on a nondigital voltmeter measure voltages up to 20.0 and 30.0 V, respectively. The resistance connected in series wi
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Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

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⇒ G=1680\ \Omega,\  G_1=2930\ \Omega

Concept to be used:

Conversion of galvanometer into voltmeter.

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To measure maximum voltage resistance R is connected in series .

So,

⇒ V=I_g(R+G)

We have to find the value of R we know that in series circuit current are same.

For G=1680                                    For G_1=2930

⇒ I_g=\frac{V}{R+G}   equation (i)                ⇒ I_g=\frac{V_1}{R+G_1} equation (ii)

Equating both the above equations:

⇒ \frac{V}{R+G} = \frac{V_1}{R+G_1}

⇒ V(R+ G_1) = V_1 (R+G)

⇒ VR+VG_1 = V_1R+V_1G

⇒ VR-V_1R = V_1G-VG_1

⇒ R(V-V_1) = V_1G-VG_1

⇒ R =\frac{V_1G-VG_1}{(V-V_1)}

⇒ Plugging the values.

⇒ R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}

⇒ R =\frac{(50400 - 58600)}{(-10)}

⇒ R=\frac{-8200}{-10}

⇒ R=820\ \Omega

The coil resistance of the circuit is 820 Ω .

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