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GenaCL600 [577]

3 years ago

7

The distance between a speaker and a listener is 8.0 m. Determine the frequency of the sound wave if exactly 20 waves are formed

before sound reaches the listener. (Speed of sound in air is 340 m/s.)

1 answer:

dmitriy555 [2]3 years ago

4 0

The frequency of the wave is 6800 Hz

<u>Explanation:</u>

Given:

Wave number, n = 20

Speed of light, v = 340 m/s

Frequency, f = ?

we know:

wave number = $\frac{frequency}{speed of light}$

$20 = \frac{f}{340 m/s} \\\\f = 20 X 340 s^-^1\\\\f = 6800 Hz$

Therefore, the frequency of the wave is 6800 Hz

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A ladder is leaning against a building so that the distance from the ground to the top of the ladder is 8 feet less than the len

SVETLANKA909090 [29]

This is the concept of application of Pythagorean theorem and algebra;
Suppose the the length of the ladder is x ft, the distance from the ground to the top of the ladder will be (x-8) ft, the distance from the bottom of the ladder to the building is 16ft, thus to find the length of the ladder we proceed as follows;
c^2=a^2+b^2
x^2=(x-8)^2+16^2
x^2=x^2-16x+64+256
collecting the like terms we get:
x^2-x^2+16x=320
16x=320
solving for x we get:
x=320/16
x=20 ft
the answer is x=20 ft

7 0

4 years ago

A tall cylinder with a cross-sectional area 12.0 cm² is partially filled with mercury; the surface of the mercury is 5.00 cm abo

JulijaS [17]

Answer:

$V = 816 cm^3$

Explanation:

As we know that gauge pressure of the fluid at the bottom of the cylinder is given as

$P = \rho g h$

now we know that pressure at the bottom is double when water is poured on the mercury

So we have

$\rho_{hg} g h_1 = \rho_w g h_2$

so we will have

$13.6 \times 5 = 1 \times h$

so we have

$h = 68 cm$

now the volume of the water added to it is given as

$V = h A$

$V = (68 cm)(12 cm^2)$

$V = 816 cm^3$

6 0

4 years ago

The power of a lens is defined as the reciprocal of its focal length: P = 1/f. (Thus power is measured in inverse meters, called

Len [333]

Answer:

This the power of combination is given by

$P_{combination}=\frac{1}{F}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}$

Explanation:

Let 'F' be the focal length of the combination of the two lenses and the focal length's of individual lenses be $f_{1},f_{2}$

We know that focal length is the position of image when object is placed at infinity

Let us place the the object at infinity with respect to the first lens thus the position of image formed by the first lens shall be obtained using lens formula as

$\frac{1}{f_{1}}=\frac{1}{u}+\frac{1}{v}$

Applying values we get

$\frac{1}{f_{1}}=\frac{1}{\infty }+\frac{1}{v}\\\\\Rightarrow \frac{1}{f}=0+\frac{1}{v}\\\\\therefore v=+f_{1}$

Now this position of image formed by the first lens act's as object for the second lens, thus we have

$\frac{1}{f_{2}}=\frac{1}{-(d-f_{1})}+\frac{1}{v}\\\\\Rightarrow \frac{1}{f_{2}}+\frac{1}{d-f_{1}}=\frac{1}{v}\\\\\therefore \frac{1}{v}=\frac{(d-f_{1})+f_{2}}{f_{2}(d-f_{1})}\\\\=\frac{1}{v}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}\\\\\therefore v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}$

Since image of an object placed at infinity will be formed at $v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}$ thus the focal length of the combination of the 2 thin lenses will be $F=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}$

This the power of combination is given by

$P_{combination}=\frac{1}{F}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}$

(For the sake of question we assume lenses to be convex although the same procedure is valid for all other lenses)

5 0

4 years ago

A 0.110 kg cube of ice (frozen water) is floating in glycerine. The glycerine is in a tall cylinder that has inside radius 3.70

Sonbull [250]

Answer:

the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

Explanation:

The change in volume of glycerin when the ice cube is placed on the surface of the glycerin can be represented as:

$V = \frac{m}{ \rho}$

Given that ;

the mass of the ice cube (m) = 0.11 kg = 0.110 × 10³ g

density of the glycerine ( $\rho$ ) = 1.260 kg/L = 1.260 g/cm³

Then:

V = $\frac{0.110*10^3 \ g}{1.260 \ g/cm^3}$

V = $0.0873*10^3 \ cm^3 (\frac{1L}{10^3 cm^3})$

V = 0.0873 L

Now;Initially the volume of the glycerin before the ice cube starts to melt is:

$V_1 = V_i + V\\\\V_1 = V_i+ 0.0873 \ L$

However; the volume of the water produced by the 0.11 kg ice cube = $0.11*10^3 \ cm^3$

The expression for change in the volume of glycerin after the ice cube starts to melt is as follows:

$V_2 = V_i + V"$

replacing $V"$ with $0.11*10^3 \ cm^3$ ; we have:

$V_2 = V_i (0.11*10^3 \ cm^3 )(\frac{1 \ L }{10^3 \ cm^3})$

$V_2 = V_i + 0.11 \ L$

The overall total change in the volume of the glycerin is illustrated as:

$V_f = V_2 - V_1$

Now; from the foregoing ; lets replace the respective value of $V_2$ and $V_1$ in the above equation ; we have;

$V_f = (V_i + 0.11 \ L) - (V_i + \ 00873 \ L)\\ \\V_f = 0.11 L - 0.0873 \ L\\\\V_f = 0.0227 \ L$

The formula usually known to be the volume of a cylinder is :

$V = \pi r ^2 h$

For the question ; we will have:

$V_f = \pi r ^2 h$

making h the subject of the formula ; we have:

$h = \frac{V_f}{\pi r^2}$

replacing 0.0227 L for $V_f$ and the given value of radius which is = 3.70 cm; we have:

$h = \frac{0.0227 \ L ( \frac{10^3 \ cm^3}{1\ L})}{\pi * (3.70 cm)^2}$

$h = \frac{22.7 \ cm^3}{\pi * (3.70 cm)^2}\\\\h = 0.528 \ \ cm$

Thus ; the distance by which the height of the liquid in the cylinder change after the ice gets melted = 0.528 cm

8 0

4 years ago

A 2,500 kg car moves at 15 m/s^2; 15m/s^2 is the (blank) of the car.

Neporo4naja [7]

A. Acceleration.
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7 0

3 years ago