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mihalych1998 [28]
3 years ago
15

Please help it’s the end the semester test

Physics
1 answer:
liq [111]3 years ago
6 0
Battery and electricity are matter ..
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P waves ________. P waves ________. are faster than S waves and surface waves have higher amplitudes than do S waves propagate o
Tomtit [17]

P waves are faster than S waves and surface waves.

<h3>Surface waves</h3>

A surface wave in physics is a mechanical wave that travels at the interface of two different media. Gravity waves on the surface of liquids, like ocean waves, are an example that is frequently used. When two liquids with different densities come together, gravity waves can also form at that junction. Solid surfaces can experience elastic surface waves like Rayleigh or Love waves. The way that electromagnetic waves can be steered along a gradient in refractive index or at the boundary between two substances with varying dielectric constants is known as "surface wave" propagation. A guided wave that travels relatively close to the Earth's surface constitutes a ground wave in radio transmission.

Learn more about surface waves here:

brainly.com/question/13779567

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5 0
2 years ago
If a fan draws 220 watts on a 120- volt circuit what is the current being drawn?
PSYCHO15rus [73]
I = 220 / 120
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4 years ago
A scuba diver has his lungs filled to half capacity (3 liters) when 10 m below the surface. If the diver holds his breath while
rjkz [21]

To solve this problem it is necessary to apply Boyle's law in which it is specified that

P_1V_1 =P_2 V_2

Where,

P_1 and V_1 are the initial pressure and volume values

P_2 and V_2 are the final pressure volume values

The final pressure here is the atmosphere, then

P_2 = 101325 \approx 1*10^5Pa

h = 10m

\rho_w = 1000kg/m^3

V_1 = 3.0L

Pressure at the water is given by,

P_1 = P_2 -\rho gh

P_1 = 1*10^5 +1000*9.8*10 =198000Pa

Using Boyle equation we have,

V_2 = \frac{P_1V_1}{P_2}

V_2 = \frac{198000*3*10^5}{10^5}

V_2 = 5.9L

Therefore the volume of the lungs at the surface is 5.9L

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3 years ago
When a force is exerted on an object, work is done only if the object
Kazeer [188]

Is moving, i'm almost positive

4 0
3 years ago
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its pla
Vikentia [17]

Answer:

The new energy density is same as initial energy density u0 as it is independent of plate dimensions

Explanation:

As we know that energy density is total energy per unit volume

so we know that isolated plates of capacitor is placed here

C = \frac{\epsilon_0 A}{d}

here we have energy density given as

u = \frac{Q^2}{2(\epsilon_0 A/d)} Ad

so we have

u = \frac{Q^2d^2}{2\epsilon_0}

so it is independent of the dimensions of the plate while total charge on the plates is constant always

so energy density will not change and remains the same

8 0
3 years ago
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