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3 years ago

The ____ is a line drawn at a right angle to a barrier. a. normal c. node b. ray d. wave front

Physics

1 answer:

Natalija [7]3 years ago

5 0

Answer:

a. normal

Explanation:

In the field of physics the normal is a line drawn at a right angle to a barrier. In other words the normal line is the line that is drawn perpendicular (right angle, 90 degrees) to the reflective surface of a mirror, or the particular boundary in which refraction occurs at the point of incidence of a light ray. This can be seen in the picture attached below.

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A rectangular block has the density of 350g/cm3 the dimensions are 3.5 6.cm 2.5

LekaFEV [45]

Answer:

18375g

Explanation:

$\boxed{density = \frac{mass}{volume} }$

$∴ mass = density \times volume$

Let's find the volume of the rectangular block.

Volume

= length ×breadth ×height

= 3.5 ×6 ×2.5

= 52.5cm³

Mass of the block

= 350(52.5)

= 18375g

7 0

2 years ago

The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplit

Lelu [443]

Answer:

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V= 30 m/s ( + x direction )

d) λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

As we know that standard form of wave equation given as

$y=A sin(\phi -\omega t)$

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

$y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s ( + x direction )

V = f λ

30 = 95.49 x λ

λ = 0.31 m

We know that speed is the rate of displacement

$U=\dfrac{dy}{dt}$

$U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]$

$U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s$

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s

8 0

3 years ago

(a) If two sound waves, one in a gas medium and one in a liquid medium, are equal in intensity, what is the ratio of the pressur

GarryVolchara [31]

Answer:

(a) The ratio of the pressure amplitude of the waves is 43.21

(b) The ratio of the intensities of the waves is 0.000535

Explanation:

Given;

density of gas, $\rho _g$ = 2.27 kg/m³

density of liquid, $\rho _l$ = 972 kg/m³

speed of sound in gas, $C_g$ = 376 m/s

speed of sound in liquid, $C_l$ = 1640 m/s

The of the sound wave is given by;

$I = \frac{P_o^2}{2 \rho C} \\\\P_o^2 = 2 \rho C I\\\\p_o = \sqrt{2 \rho CI}$

Where;

$P_o$ is the pressure amplitude

$P_o_g= \sqrt{2 \rho _g C_gI} -------(1)\\\\P_o_l= \sqrt{2 \rho _l C_lI}---------(2)\\\\\frac{P_o_l}{P_o_g} = \frac{\sqrt{2 \rho _l C_lI}}{\sqrt{2 \rho _g C_gI}} \\\\\frac{P_o_l}{P_o_g} = \sqrt{\frac{2 \rho _l C_lI}{2 \rho _g C_gI} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ \rho _l C_l}{ \rho _g C_g} }\\\\ \frac{P_o_l}{P_o_g} = \sqrt{\frac{ (972)( 1640)}{ (2.27)( 376)} }\\\\\frac{P_o_l}{P_o_g} = 43.21$

(b) when the pressure amplitudes are equal, the ratio of the intensities is given as;

$I = \frac{P_o^2}{2 \rho C}\\\\I_g = \frac{P_o^2}{2 \rho _g C_g}-------(1)\\\\I_l = \frac{P_o^2}{2 \rho _l C_l}-------(2)\\\\\frac{I_l}{I_g} = (\frac{P_o^2}{2 \rho _l C_l})*(\frac{2\rho_gC_g}{P_o^2} )\\\\\frac{I_l}{I_g} = \frac{\rho _gC_g}{\rho_lC_l} \\\\\frac{I_l}{I_g} = \frac{(2.27)(376)}{(972)(1640)}\\\\ \frac{I_l}{I_g} = 0.000535$

3 0

3 years ago

What is an non example of Cultural diffusion

belka [17]

Beliefs is a non example

4 0

3 years ago

One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.

Zepler [3.9K]

Answer:

The answer is " $\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}$ "

Explanation:

Given:

$\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\$

calculation:

$Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978$

4 0

2 years ago