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arsen [322]
3 years ago
11

How does Earth's surface and the structures on the surface change as a result of an earthquake?

Physics
1 answer:
olga2289 [7]3 years ago
8 0

THE CHANGE IN EARTH SURFACE IS DUE TO VIBRATION AND COLLISION IN EARTH TECTONIC PLATES

Explanation:

  • VIBRATION is the change in the stable state/equilibrium state caused by increased pressure in a particular object or a surface due to electro magnetic wave
  • COLLISION is the meeting point of two or more surfaces or the objects
  • During earthquake the tectonic plates of the earth surface begins to collide or cause any irregular drastic movements
  • This is due to the vibration caused by earthquake
  • This causes the shake in the upper surface which results in construction damage and calamities
  • This causes drastic change in the surface of the earth.
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Why are temperatures of the currents generally colder at the poles than the equator?
patriot [66]

Answer:

the answer is A

Explanation:

6 0
3 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

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

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

7 0
3 years ago
Two everyday examples of a starting motion
Tasya [4]
Pushing, pulling is the answer
8 0
3 years ago
A e B são dos blocos de massas 3,0 kg e 2,0 kg, respectivamente, que se movimentam juntos sobre uma superficie horizontal e perf
makvit [3.9K]

F = m*a

30 N = (ma + mb) * a

30 = 5*a

a = 6 m/s ^2

F de B em A

30 - F de B,A = ma * a

30 - F de B em A = 3 * 6

30 - 18 = F de B em A

12 = F de B em A


Resposta: 6 m/s^2 e 12N

Bate com o gabarito, man? Ou eu tô viajando aqui?

Abç!

6 0
3 years ago
Two loudspeakers in a 20c room emit 686 hz sound waves along the x-axis.a. if the speakers are in phase, what is the smallest di
jeyben [28]

<u>Answer :</u>

(a) d = 0.25 m

(b) d = 0.5 m

<u>Explanation :</u>

It is given that,

Frequency of sound waves, f = 686 Hz

Speed of sound wave at 20^0\ C is, v = 343 m/s

(1) Perfectly destructive interference occurs when the path difference is half integral multiple of wavelength i.e.

d=\dfrac{\lambda}{2}........(1)

Velocity of sound wave is given by :

v=f\times \lambda

d=\dfrac{v}{2f}

d=\dfrac{343\ m/s}{2\times 686\ Hz}

d=0.25\ m

Hence, when the speakers are in phase the smallest distance between the speakers for which the interference of the sound waves is perfectly destructive is 0.25 m.

(2) For constructive interference, the path difference is integral multiple of wavelengths i.e.

d=n\lambda  ( n = integers )

Let n = 1

So, d=\dfrac{v}{f}

d=\dfrac{343\ m/s}{686\ Hz}

d=0.5\ m

Hence, the smallest distance between the speakers for which the interference of the sound waves is maximum constructive is 0.5 m.

4 0
3 years ago
Read 2 more answers
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