All categories

3 years ago

If the wave is detected 12.5 minutes after the earthquake, estimate the distance from the detector to the site of the quake

Physics

1 answer:

umka2103 [35]3 years ago

5 0

Answer:

Remember the relation:

Speed*Time = Distance.

We can estimate that the speed at which an earthquake "moves", in the surface, is:

S = 6km/s (this is a low estimation actually)

Then if the wave is detected 12.5 minutes after the earthquake, we know that it traveled for 12.5 minutes before reaching the detector.

So we know the speed of the wave and the time it took to reach the detector, then we can use the equation:

Speed*Time = Distance.

to find the distance.

First, we should write the time in seconds

we know that:

1 min = 60 s

then:

12.5 min = 12.5*(60 s) = 750 s

Then, the wave traveled with a speed of 6 km/s for 750 seconds until it reached the detector, then the distance that it traveled is:

(6km/s)*750s = 4500 km

The distance between the detector and the site of the quake is around 4500 km.

You might be interested in

Anybody know these ?

pashok25 [27]

Answer:

1. a) 72 N.

2. a) 2 m/s².

Explanation:

Given the following data;

1. Mass = 90kg

Acceleration = 0.8 m/s²

To find the force;

Force = mass * acceleration

Force = 90 * 0.8

Force = 72 Newton.

2. Mass = 50kg

Force = 100N

To find the magnitude of acceleration;

Acceleration = force/mass

Acceleration = 100/50

Acceleration = 2 m/s²

5 0

3 years ago

Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

3 years ago

Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.

n200080 [17]

Answer:

The answer to your question is the letter A) F = 9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

F = k q₁q₂ /r²

-Substitution

F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

F = 3.512 x 10⁻⁷ / 0.381

-Result

F = 9.227 x 10⁻⁷ N ≈ 9.23 x 10⁻⁷ N

8 0

3 years ago

All Physicsts over here plz help in these questions!!!!!!!!!

mrs_skeptik [129]

Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.

6 0

3 years ago

Read 2 more answers

Please don't post anything unless it's the answer

Alja [10]

52800000000000000000000000000000000000000000

7 0

3 years ago

If the wave is detected 12.5 minutes after the earthquake, estimate the distance from the detector to the site of the quake​

If the wave is detected 12.5 minutes after the earthquake, estimate the distance from the detector to the site of the quake