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3 years ago

How can you keep sound out of a room

Physics

1 answer:

GREYUIT [131]3 years ago

8 0

To keep<span> noise from entering your space, look for </span>sound<span> blockers</span>

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A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi

ella [17]

Answer:

$\mu=1.5322$

Explanation:

Given:

Thickness of the paperweight cube, $x=8.55\ cm$

apparent depth from one side of the inbuilt paper in the plastic cube, $i=4.02\ cm$

apparent depth from the other side of the inbuilt paper in the plastic cube, $i'=1.56\ cm$

Now as we know that refractive index is given as:

$\rm \mu=\frac{real\ depth}{apparent\ depth}$

Let the real depth form first side of the slab be, $d$

Then the depth from the second side of the slab will be, $d'=x-d=8.55-d$

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

$\mu=\mu'$

$\frac{d}{i} =\frac{d'}{i'}$

$\frac{d}{4.02}=\frac{8.55-d}{1.56}$

$d=6.1597\ cm$

Now the refractive index:

$\mu=\frac{d}{i}$

$\mu=\frac{6.1596}{4.02}$

$\mu=1.5322$

7 0

3 years ago

What energy transformation occurs when a skydiver first jumps from a plane

d1i1m1o1n [39]

The skydiver jumping from a plane high up in the sky would most likely experience various energy transformation. For starters, it would undergo a very large gravitational potential energy because of its much higher elevation. After jumping, this energy would eventually transform to kinetic energy due to the force exerted by the gravity.

3 0

3 years ago

Read 2 more answers

Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in and the atom emits a photon of lig

PSYCHO15rus [73]

Give me some answer choices and i will be happy to help

4 0

4 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$