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pantera1 [17]
3 years ago
6

In what direction do seismic waves carry the energy of an earthquake?

Physics
2 answers:
irinina [24]3 years ago
8 0
Seismic waves carry energy from an earthquake away from the focus, through earth's interior and across the surface.
Oxana [17]3 years ago
3 0
<h3><u>Answer;</u></h3>

<em>Away from the focus</em>

Seismic waves carry energy from an earthquake <em><u>away from the focus</u></em>, through earth's interior and across the surface.

<h3><u>Explanation</u>;</h3>
  • <em><u>When an earthquake occurs energy is produced and is transmitted or released in form of a seismic waves.</u></em> Seismic waves are types of waves which results from an earthquake,<em><u> they are vibrations that move through the earth carrying energy</u></em> that has been released when an earthquake occurs.
  • The <u><em>focus is the point at which a rock that is under stress breaks and triggers and earthquake</em></u>. Seismic waves carry the energy released during an earthquake<em><u> away from the focus</u></em> through the earth in all directions.
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Three Small Identical Balls Have Charges -3 Times 10^-12 C, 8 Times 10^12 C And 4 Times 10^-12 C Respectively. They Are Brought
IgorLugansk [536]

Answer:

The charge in each ball will be 3 * 10^-12 C

Explanation:

(Assuming the correct charge of the second ball is 8 * 10^-12)

When the balls are brought in contact, all the charges are split evenly among then.

So first we need to find the total charge combined:

(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C

Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:

(9 * 10^-12) / 3 = 3 * 10^-12 C

So the charge in each ball will be 3 * 10^-12 C

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3 years ago
Often times we think about insulation, we think about keeping cold items cold, such as drinks in an ice chest or the cool temper
Burka [1]

Answer:

So, insulation essentially works by creating a sort of barrier between the hot and the cold object. This barrier helps to reduce heat transfer by either reflecting the thermal radiation or by decreasing thermal conduction and convection from one object to the other.

4 0
2 years ago
Suppose a capacitor is fully charged by a battery and then disconnected from the battery. The positive plate has a charge +q and
dimaraw [331]

Answer:-q

Explanation:

Given

Capacitor is charged to a battery and capacitor acquired a charge of q i.e.

+q on Positive Plate and -q on negative Plate.

If the plate area is doubled and the plate separation is reduced to half its initial separation then capacitor becomes four times of initial value because capacitor is given by

c=\epsilon_0 \cdot \frac{A}{d}

where A=area of capacitor plate

d=Separation between plates

This change in capacitance changes the Potential such that new charge on the negative plate will remain same -q

8 0
3 years ago
A circular coil of radius r = 5 cm and resistance R = 0.2 is placed in a uniform magnetic field perpendicular to the plane of th
Yuri [45]

Answer:

the question is incomplete, the complete question is

"A circular coil of radius r = 5 cm and resistance R = 0.2 ? is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.5 e^-t T. What is the magnitude of the current induced in the coil at the time t = 2 s?"

2.6mA

Explanation:

we need to determine the emf induced in the coil and y applying ohm's law we determine the current induced.

using the formula be low,

E=-\frac{d}{dt}(BACOS\alpha )\\

where B is the magnitude of the field and A is the area of the circular coil.

First, let determine the area using \pi r^{2} \\ where r is the radius of 5cm or 0.05m

A=\pi *(0.05)^{2}\\ A=0.00785m^{2}\\

since we no that the angle is at 0^{0}

we determine the magnitude of the magnetic filed

B=0.5e^{-t} \\t=2s

E=-(0.5e^{-2} * 0.00785)

E=-0.000532v\\

the Magnitude of the voltage is 0.000532V

Next we determine the current using ohm's law

V=IR\\R=0.2\\I=\frac{0.000532}{0.2} \\I=0.0026A

I=2.6mA

6 0
4 years ago
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Explanation:

It is given that,

Length of the helicopter, l = 3.1 m

The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m

Angular speed of the helicopter, \omega=280\ rev/min=29.32\ rad/s

(a) The centripetal acceleration in terms of angular velocity is given by :

a_c=r\times \omega^2

a_c=3.1\times (29.32)^2

a_c=2664.95\ m/s^2

(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :

v=r\times \omega

v=3.1\times 29.32

v = 90.89 m/s

\dfrac{v_{tip}}{v_{sound}}=\dfrac{90.89}{340}=0.267

Hence, this is the required solution.

4 0
3 years ago
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