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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

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

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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.

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4 years ago