All categories

3 years ago

Why is a minimum of three seismic stations needed to find the epicenter of an earthquake?

1 answer:

5 0

Each station can detect how far away the epicenter was. So each station basically has a circle made of possible epicenters. When you have three, you narrow it down to one, final point.

You might be interested in

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

Nuclear sizes are expressed in a unit named

o-na [289]

Answer:

Answer is A) Fermi

Explanation:

Fermi is the expressive unit for nuclear sizes. Fermi = 10^-15 meter.

4 0

3 years ago

How much energy is needed to perform 300 J of work?

4vir4ik [10]

I think this is the answer hope it helps

3 0

3 years ago

WHAT SHOULD I NAME MY DEAD RAT?

olga_2 [115]

Answer:

pablito

Explanation:

6 0

3 years ago

A gyroscope flywheel of radius 3.25 cm is accelerated from rest at 11.6 rad/s2 until its angular speed is 1820 rev/min. (a) What

kati45 [8]

Explanation:

The given data is as follows.

radius (r) = 3.25 cm, $\alpha = 11.6 rad/s^{2}$

Now, we will calculate the tangential acceleration as follows.

$a_{tangential} = \alpha \times r$

Putting the given values into the above formula as follows.

$a_{tangential} = \alpha \times r$

= $11.6 rad/s^{2} \times 3.25 cm$

= 37.7 $rad cm/s^{2}$

Thus, we can conclude that the tangential acceleration of a point on the rim of the flywheel during this spin-up process is 37.7 $rad cm/s^{2}$ .

6 0

3 years ago