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Verizon [17]
3 years ago
14

A tightened string vibrates with a standing wave. Which of the following statements is correct? Group of answer choices Points o

n the string undergo the same displacement. Points on the string vibrates with different amplitude. Points on the string vibrates with different frequencies. Points on the string undergo the same speed. Points on the string vibrate with the same energy..
Physics
1 answer:
jek_recluse [69]3 years ago
6 0

Nodes and antinodes on a vibrating string standing wave has different amplitudes.

Option B

<u> Explanation: </u>

  • For a standing wave produced from a tightened string vibration, points of maximum and minimum displacement are formed termed as nodes and antinodes. Hence (A) is false.
  • The amplitudes at nodes are zero and increases gradually to the maximum when it approaches the antinode. Statement (B) is true.
  • The string vibrates with a single natural frequency and a number of resonant frequencies. The nodes and antinodes have same frequencies. (C) is false.
  • Velocity at the nodes is zero and it increases gradually at the antinodes. Statement (D) is false.
  • Since nodes have a displacement equal to zero, no energy is present. All energy is confined in between two nodes. Energy at any point on the string is always constant and is not transferred.(E) is false.  
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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a
makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°

The distance r is:

r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m

You replace the values of all parameters in the equation (1):

\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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