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3 years ago

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>Answer;</h3>

Away from the focus

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

<h3>Explanation;</h3>

When an earthquake occurs energy is produced and is transmitted or released in form of a seismic waves. Seismic waves are types of waves which results from an earthquake, they are vibrations that move through the earth carrying energy that has been released when an earthquake occurs.
The focus is the point at which a rock that is under stress breaks and triggers and earthquake. Seismic waves carry the energy released during an earthquake away from the focus through the earth in all directions.

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What to do when having a cold

BigorU [14]

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Explanation:

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3 years ago

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3 years ago

4 points

zubka84 [21]

(a) 25lx

(b) 11.11lx

Explanation:

Illuminance is inversely proportional to the square of the distance.

So,

$I = k\frac{1}{r^2}$

where, k is a constant

So,

(a)

If I = 100lx and r₂ = 2r Then,

$I_2 = k\frac{1}{(2r)^2}$

Dividing both the equation we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(2r)^2}{k} \\\\\frac{I_1}{I_2} = 4\\\\I_2 = \frac{I_1}{4}\\\\I_2 = \frac{100}{4} = 25lx$

When the distance is doubled then the illumination reduces by one- fourth and becomes 25lx

(b)

If I = 100lx and r₂ = 3r Then,

$I_2 = k\frac{1}{(3r)^2}$

Dividing equation 1 and 3 we get

$\frac{I_1}{I_2} = \frac{k}{r^2} X\frac{(3r)^2}{k} \\\\\frac{I_1}{I_2} = 9\\\\I_2 = \frac{I_1}{9}\\\\I_2 = \frac{100}{9} = 11.11lx$

When the distance is tripled then the illumination reduces by one- ninth and becomes 11.11lx

3 0

3 years ago

According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something

Tasya [4]

Answer:

The pressure is $p_1 = 4051.4 \ Pa$

Explanation:

From the question we are told that

The gauge pressure at the mouth is $p_1$

The radius of the column is $r_2 = 4 \ mm = 0.004 \ m$

The speed of the liquid outside the body is $v_2 = 3.1 \ m/s$

The area of the column is $A_2$

The area inside the mouth $A_1 = 10 A_2$

Generally according to continuity equation

$v_1 A_1 = v_2 A_2$

=> $v_ 1 = v_2 * \frac{A_2}{A_1}$

=> $v_ 1 = 3.1 * \frac{1}{10}$

=> $v_ 1 = 0.31 \ m/s$

$A_1 = 10A_2$

=> $\pi * r_1^2 = 10(\pi * r_2^2)$

=> $r_1 = 10 * r_2$

substituting values

$r_1 = 10 * 0.004$

$r_1 =0.04 \ m$

Now the height of inside the mouth is $h_1 = d = 2r_1 = 2* 0.04 = 0.08\ m$

Now the height of the column is $h_2 = d = 2r_2 = 2* 0.004 = 0.008\ m$

Generally according to Bernoulli's equation

$p_1 = [\frac{1}{2} \rho v_2^2 + h_2 \rho g +p_2] -[\frac{1}{2} \rho * v_1^2 + h_1 \rho g ]$

Now $\rho = 1000 \ kg m^{-3}$ which is the density of water

$p_2$ is the gauge pressure of the atmosphere which is zero

$p_1 = [(0.5 * 1000 * (3.1)^2) +(0.008 * 1000 * 9.8) + 0]-$

$[(0.5 * 1000 * 0.31^2) + (0.08*1000 * 9.8)]$

$p_1 = 4051.4 \ Pa$

8 0

3 years ago

A car was at a 30km marker and began to drive. The car passed a 90km marker and the time had changed 30

dolphi86 [110]

Answer:

i think it's 2km pm

Explanation:

2km x 30 60.. start was 30, and now your at 90.. we had to determine how much time it took.. so 2 is the average.. or atleast per minute and sorry it i still didnt answer ur question lol im just trynna help

7 0

2 years ago