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3 years ago

What wave travels through water and air?????

Physics

2 answers:

ss7ja [257]3 years ago

8 0

I think the answer is sound

finlep [7]3 years ago

7 0

Primary Waves
Secondary go only through air, but not water.

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How do I do this?<br> Series and parallel circuits.

Svetlanka [38]

Connect the little light thingys to the top of the battery

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3 years ago

What is the rotational inertia of an object that rolls without slipping down a 2.00-m-high incline starting from rest, and has a

krek1111 [17]

We will apply the concepts of energy conservation to solve the problem. We know that gravitational potential energy is equivalent to the sum of translational kinetic energy and rotational kinetic energy. Additionally, the relation of the angular velocity with the tangential velocity will be determined to eliminate the angular term and obtain the expression of the Inertia in terms of the data given, therefore, we know that

$PE_{g} = KE_{trans} +KE_{rot}$

$mgh = \frac{1}{2} mv^2+\frac{1}{2}I\omega^2$

The angular velocity in terms of tangential velocity and radius is defined as,

$\omega = \frac{v}{R}$

Replacing,

$mgh = \frac{1}{2} mv^2+\frac{1}{2} I(\frac{v}{R})^2$

Multiplying by R^2,

$gh(mR^2) = \frac{v^2}{2} (mR^2)+\frac{v^2}{2}I$

$\frac{v^2}{2}I = (gh-\frac{v^2}{2})mR^2$

$I = (\frac{2gh}{v^2}-1)mR^2$

Replacing with our values we have,

$I = (\frac{2(9.8)(2)}{6^2}-1)mR^2$

$I = 0.089mR^2$

6 0

3 years ago

Can someone help me with this. I'm not really sure if the right answer is c.

taurus [48]

It might be c I'm not sure either. Hope I helped

6 0

3 years ago

What is the difference between current in series and in parallel

Hatshy [7]

Answer:

In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component. ... In a series circuit, every device must function for the circuit to be complete. If one bulb burns out in a series circuit, the entire circuit is broken.

3 0

3 years ago

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago