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Olin [163]
4 years ago
11

Two similar waves X and Y travel through the same medium. Wave X has a frequency of 200 hertz and a

Physics
1 answer:
velikii [3]4 years ago
4 0

Answer:

The correct solution is option A "10 meters".

Explanation:

The given values are:

Wave X,

Frequency,

F_x=200 \ H_z

Wavelength

\lambda_x=35 \ m

Wave Y,

Frequency,

F_y=700 \ Hz

Let its wavelength = \lambda_y

As we know,

⇒  v=f \lambda

For both waves, medium is same then

⇒  v_x=v_y

⇒  f_x \lambda_x=f_y \lambda_y

then,

⇒  \lambda_y=\frac{f_x \lambda_x}{f_y}

On substituting the estimated values in the above equation, we get

⇒      =\frac{200\times 35}{700}

⇒      =\frac{7000}{700}

⇒      =10 \ m

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The next two questions refer to this situation: A rectangular loop with sides of length a= 1.00 cm and b= 2.70 cm is placed near
alisha [4.7K]

Answer:

a) 1.007 * 10^{-7} V.m , into the page

b) -5.39 * 10^{-8} V, counterclockwise

Explanation:

a) d\Phi = B.dA\\\\B_r = \frac{\mu_0I}{2\pi r}\\\\dA = b.dr\\\\\Phi = \int\limits^b_a {B.dA} = \int\limits^{a+d}_d {\frac{\mu_0I}{2\pi r}.b.dr} =\frac{\mu_0I}{2\pi }.b.ln(\frac{a+d}{d} )\\

At t = 2.90, I = 3.62 + 1.49 * (2.90)^2 = 16.15 A

\Phi = \frac{4\pi * 10^{-7}*16.15*0.027}{2\pi} ln(1.46/0.46) = 1.007 * 10^{-7} V.m

Direction is into the page.

b) emf = -d\Phi/dt = -\frac{\mu_0}{2\pi }.b.ln(\frac{a+d}{d}).(2.98t)= \\=-\frac{4\pi * 10^{-7}*0.027}{2\pi} ln(1.46/0.46)*2.98*2.90= -5.39 * 10^{-8} V

Direction is counterclockwise.

8 0
3 years ago
A dog in an open field runs 11.0 m east and then 28.0 m in a direction 49.0 ∘ west of north. In what direction must the dog then
Ede4ka [16]

Answer:

\theta=19.03 SE

Explanation:

From the question we are told that:

d_1=11.0m East

d_2=28.0m 49 \textdegree west\ north

d_3=10m south

Generally the equation for Resolutions is mathematically given by

d_x=-11-(-28sin49)

d_x=10.132m

Where

d_y=-11-(28cos49)

d_y=-29.37m

Generally the equation for Direction is mathematically given by

\theta=tan^{-1}(\frac{dx}{dy})

\theta=tan^{-1}(\frac{10.132}{29.37})

\theta=19.03 SE

7 0
3 years ago
The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and
mina [271]

Answer:

\frac{dQ}{dt}= 4312 W

Explanation:

As we know that base of the slab is given as

A = 11 \times 8

A = 88 m^2

now we know that rate of heat transfer is given as

\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)

here we know that

k = 1.4 W/m k

Also we have

x =0.20

\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)

\frac{dQ}{dt}= 4312 W

7 0
3 years ago
Q11:A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5
natulia [17]

Answer:

   v_s = 34.269 m / s

Explanation:

This is a Doppler effect exercise, in this case the observer is fixed and the sound source is moving.

         f ’= f  \frac{v}{v  \mp v_s  }

where the negative sign is used for when the source approaches the observer and the positive sign for when the source moves away from the observer

In this case when f ’= 5500 Hz approaches and when f’ = 4500 Hz moves away, let's write the two expressions together

          5500 = f (\frac{v}{v - v_s })

          4500 = f ( \frac{v}{v + vs})

let's solve these two equations

           \frac{5500}{4500}     = \frac{v+v_s}{v-v_s}

           1.222 (v-v_s) = v + v_s

            v_s (1+ 1.22) = v (1.222 -1)

            v_s = v   \frac{0.222}{2.223}

the speed of sound in air is v = 343 m / s

            v_s = 343 0.09990

 

             v_s = 34.269 m / s

4 0
3 years ago
A body has a velocity of 143 km/hr. Find its value in m/s.
kozerog [31]
39.72 m/s is the velocity of the body.

3 0
4 years ago
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