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3 years ago

The wave y1 = 0.12 sin π/3(5x+200t-1) propagates on a string of linear density 0.02 kg/m.

1 answer:

5 0

The average power transmitted by string wave is given by the expression $P=\frac{1}{2} \mu\omega^2A^2v$ , Where μ is a linear density of a string, ω is angular frequency of the wave, A is an amplitude of the wave, v is a speed of wave.

Since $y(x,t)= A sin (kx+\omega t-\psi )$ comparing with $y_1 = 0.12 sin \frac{\pi}{3}(5x+200t-1)$

A = 0.12 m, ω = $\frac{200\pi }{3} s^{-1}$ , k = $\frac{5\pi }{3} m^{-1}$

Speed of wave, $v = \frac{\omega }{k} = \frac{\frac{200\pi }{3}}{\frac{5\pi }{3} } =40m/s$

So power $P = \frac{1}{2} *0.02*(\frac{200\pi }{3} )^2*0.12^2*40=253 W$

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A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

HACTEHA [7]

Answer:

$B=2.91\ \mu T$

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$

Put all the values,

$B=10^{-7}\times \dfrac{2\pi \times 0.4^2 \times 2}{(0.4^2+0.09^2)^{3/2}}\\\\=2.91\times 10^{-6}\ T\\\\or\\\\B=2.91\ \mu T$

So, the required magnetic field is equal to $2.91\ \mu T$ .

3 0

3 years ago

I pooped in the bathtub? I tryed to poop in my hand so i can carry it to the fish tank so they can feed thereselfs but it fell t

cupoosta [38]

Answer:

Sorry don't know

Explanation:

Ask google

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3 years ago

Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T

ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

$\vec{A}=x_A\hat{i}$

Vector B , which has a magnitude of 5.5 m

$\vec{B}=x_B\hat{i}+y_B\hat{j}$

$\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25$

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A $\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}$