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3 years ago

Compare and contrast a transverse wave and a compressional wave Give an example for each type

Physics

1 answer:

kari74 [83]3 years ago

6 0

Answer:

Transverse waves oscillate perpendicular to the direction of the wave (e.g. any electromagnetic wave like radiowaves, x-rays...) whilst compressional waves oscillate in the same direction of the wave (e.g. sound waves)

Explanation:

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A parallel-plate capacitor has plates of area A. The plates are initially separated by a distance d, but this distance can be va

ohaa [14]

Answer:

d' = d /2

Explanation:

Given that

Distance = d

Voltage =V

We know that energy in capacitor given as

$U=\dfrac{1}{2}CV^2$

$C=\dfrac{\varepsilon _oA}{d}$

$U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2$

If energy become double U' = 2 U then d'

$U'=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2U=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

$2\times \dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d}\times V^2=\dfrac{1}{2}\times \dfrac{\varepsilon _oA}{d'}\times V^2$

2 d ' = d

d' = d /2

So the distance between plates will be half on initial distance.

7 0

3 years ago

3. A rocket is launched at an angle of 53 degrees above the 1 point

irina1246 [14]

Answer:

24,000 m

Explanation:

First find the rocket's final position and velocity during the first phase in the y direction.

Given:

v₀ = 75 sin 53° m/s

t = 25 s

a = 25 sin 53° m/s²

Find: Δy and v

Δy = v₀ t + ½ at²

Δy = (75 sin 53° m/s) (25 s) + ½ (25 sin 53° m/s²) (25 s)²

Δy = 7736.8 m

v = at + v₀

v = (25 sin 53° m/s²) (25 s) + (75 sin 53° m/s)

v = 559.0 m/s

Next, find the final position of the rocket during the second phase (as a projectile).

Given:

v₀ = 559.0 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (559.0 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 15945.5 m

The total displacement is:

7736.8 m + 15945.5 m

23682.2 m

Rounded to two significant figures, the maximum altitude reached is 24,000 m.

3 0

3 years ago

Calculate the illuminance at sea level from the sun at zenith, using the following information: The luminance of the sun at zeni

Sergeeva-Olga [200]

Answer:

346.01 × 10² Lux

Explanation:

Given:

luminance of the sun at zenith at sea level, Ls = 1600 × 10 cd/m²

The diameter of the sun's photosphere = 8.64 × 10 miles = 45.62 × 10⁸ ft

Radius, r = $\frac{45.62\times10^8\ ft}{\textup{2}}$

r = 22.81 × 10⁸ ft

The distance from the sun to the earth = 92.9 × 10 miles = 49.05 x 10¹⁰ ft

Now,

Lumen = Luminance × 4πr²

Lumen = 1600 × 10 cd/m² × 4πr² .....................(1)

also,

Illumination = $\frac{\textup{Lumen}}{\textup{4}\pi\textup{Distance}^2}$

on substituting lumen from 1

Illumination = $\frac{1600\times10\times4\pi r^2}{\textup{4}\pi\textup{Distance}^2}$

Illumination = $\frac{1600\times10^6\times4\pi\times(22.81\times10^8\ ft)^2}{\textup{4}\pi\textup{49.05\times10^10 ft}^2}$

Illumination = 346.01 × 10² Lux

5 0

4 years ago

A chemical reaction that has the general formula of AB + C → CB + A is best classified as a reaction.

MissTica

Answer:

Single replacement

Explanation:

A reaction in which one element replaces a similar element is called single replacement. In this case, C is replacing A.

7 0

4 years ago

Read 2 more answers

Help me with this problem please

zaharov [31]

Answer:

Total moment of inertia when arms are extended: 1.613 $kg\,m^2$

Explanation:

This second part of the problem could be a pretty complex one, but if they expect you to do a simple calculation, which is what I imagine, the idea is just adding another moment of inertia to the first one due to the arms extended laterally and use the moment of inertia for such as depicted in the image I am attaching.

In that image:

L is the length from one end to the other of the extended arms (each 0.75m from the center of the body) which gives 1.5 meters.

m is the mass of both arms. That is: twice 5% of the mass of the person: which mathematically can be written as: 2 * 0.05 * 56.5 = 5.65 kg

Therefore this moment of inertia to be added can be obtained using the formula shown in the image:

$I_z=\frac{1}{12} \,m\,L^2\\\\I_z=\frac{5.65\,*\,1.5^2}{12} \\I_z=1.05937\,kg\,m^2$

Now, one needs to add this to the previous moment that you calculated, resulting in:

0.554 + 1.059 = 1.613 $kg\,m^2$

5 0

3 years ago