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3 years ago

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To navigate, a porpoise emits a sound wave that has a wavelength of 4.9 cm. The speed at which the wave travels in seawater is 1

522 m/s. Find the period of the wave.

1 answer:

bekas [8.4K]3 years ago

5 0

Frequency can be defined as the ratio between the speed of the wave and its wavelength, that is

$f= \frac{v}{\lambda}$

At the same time, the frequency is the inverse of the Period so

$T = \frac{1}{f}$

If we join the two expressions we will have to

$T = \frac{\lambda}{v}$

Replacing we have that

$T = \frac{4.9*10^{-2}}{1522}$

$T = 3.219*10^{-5} s$

Therefore the period of the wave is $3.219*10^{-5} s$

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A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate

babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$

3 0

3 years ago

Most of the stars in the Milky Way will end their lives as

umka21 [38]

<span> </span>Most of the stars in the Milky Way will end their lives as white dwarfs

4 0

2 years ago

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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

pishuonlain [190]

Answer:

$E_{total}=4.82*10^6N/C$

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

$E_{total}=E_1+E_2$

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

$E=\sigma/(2\epsilon_o)$

Then:

$E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C$

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0

3 years ago

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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

$S_1 S_2$ =3m

$S_1 O$ =4m

By Pythagoras theorem

$S_2O=\sqrt{(S_1S_2)^2+(S_1O)^2}\\S_2O=\sqrt{(3)^2+(4)^2}\\S_2O=\sqrt{9+16}\\S_2O=\sqrt{25}\\S_2O=5m$

Now

The intensity at O when both speakers are on is given by

$I=4I_1 cos^2(\pi \frac{\delta}{\lambda})$

Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

$16=4\times 6 \times cos^2(\pi \frac{1 \times f}{320})\\16/24= cos^2(\pi \frac{1f}{320})\\0.667= cos^2(\pi \frac{f}{320})\\cos(\pi \frac{f}{320})=\pm0.8165\\\pi \frac{f}{320}=\frac{7 \pi}{36}+2k\pi \\ \frac{f}{320}=\frac{7 }{36}+2k \\\\ {f}=320 \times (\frac{7 }{36}+2k )$

where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

So the minimum frequency is 702.22 Hz

5 0

2 years ago

A 8.34 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is

Mnenie [13.5K]

Answer:

21870.3156 N

Explanation:

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 1.6 m/s²

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-18.2^2}{2\times 162}\\\Rightarrow a=-1.02234\ m/s^2$

The acceleration of the craft should be 1.02234 m/s²

$F=ma\\\Rightarrow F=8.34\times 10^3\times 1.02234\\\Rightarrow F=8526.3156\ N$

Weight of the craft

$W=mg\\\Rightarrow W=8.34\times 10^3\times 1.6\\\Rightarrow W=13344\ N$

Thrust

$F_t=F+W\\\Rightarrow F_t=8526.3156+13344\\\Rightarrow F_t=21870.3156\ N$

The thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface is 21870.3156 N

3 0

3 years ago