Ask question

Ask question

All categories

3 years ago

6

What is the application for reflection of waves?

1 answer:

Kazeer [188]3 years ago

5 0

One big application is the manufacture and use of mirrors.

You might be interested in

Water ice has a density of 0.91 g/cm2, so it will float in liquid water. Imagine you have a cube of ice, 10 cm on a side. a. Wha

Reptile [31]

Answer:

(i) W = 8.918 N

(ii) $V = 9.1 \times 10^{-4} m^3$

(iii) d = 9.1 cm

Explanation:

Part a)

As we know that weight of cube is given as

$W = mg$

$W = \rho V g$

here we know that

$\rho = 0.91 g/cm^3$

$Volume = L^3$

$Volume = 10^3 = 1000 cm^3$

now the mass of the ice cube is given as

$m = 0.91 \times 1000 = 910 g$

now weight is given as

$W = 0.910 \times 9.8 = 8.918 N$

Part b)

Weight of the liquid displaced must be equal to weight of the ice cube

Because as we know that force of buoyancy = weight of the of the liquid displaced

$W_{displaced} = 8.918 N$

So here volume displaced is given as

$\rho_{water}Vg = 8.918$

$1000(V)9.8 = 8.918$

$V = 9.1 \times 10^{-4} m^3$

Part c)

Let the cube is submerged by distance "d" inside water

So here displaced water weight is given as

$W = \rho_{water} (L^2 d) g$

$8.918 = 1000(0.10^2 \times d) 9.8$

$d = 0.091 m$

so it is submerged by d = 9.1 cm inside water

4 0

4 years ago

Which has the same meaning as air pressure?

SVETLANKA909090 [29]

B) atmosphere pressure, i believe

7 0

3 years ago

A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continu

kolbaska11 [484]

Answer:

τ=0.060 N.m

Explanation:

By kinematics:

$\omega f = \omega o-\alpha*t$

Solving for α:

$\alpha=\frac{\omega o-\omega f}{t}$

where ωo = 600*2*π/60; ωf = 0; t=10s

$\alpha=6.283rad/s^2$

The sum of torque is:

$\tau=I*\alpha$

$\tau=M*R^2/2*\alpha$

$\tau=0.060 N.m$

8 0

3 years ago

1) Think of planting a tree. You dig the hole, put it in, and fill it back in. Next,

pogonyaev

Answer:

at the top of the tree I hope it will help you please follow me

6 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago