All categories

3 years ago

4. Why is it better to use data from three or more seismic stations to find the epicenter of an earthquake?

2 answers:

6 0

Ilana is togladly hospitalized for the loss and the loss and her family of the dead were in a coma and she had to go on the plane and then to go to a hotel and to see if the girl had a child and a woman was in a car crash or something like it could have caused a lot to happen with her body that could help him get out the next morning after a long day of work and a woman was in a condition of a new yo girl in her apartment and was taken into a car in a coma at a

Umnica [9.8K]3 years ago

5 0

Answer:Data collected from one seismic station can only determine the distance from the epicentre to this station, thus producing a circle with a radius of that distance. If we have two stations, there will be two circles, and these two circles will have two intercepting points (can't determine the exact location). Therefore we need to add another circle, hence another station to know the exact location of the epicentre.

Explanation:

You might be interested in

A solid sphere has a temperature of 614 K. The sphere is melted down and recast into a cube that has the same emissivity and emi

krok68 [10]

Answer:581.87 K

Explanation:

Given

Sphere is melted to form a square

Let the radius of sphere be r and square has a side a

Therefore

$\frac{4\pi}{3}r^3=a^3$

Surface area of sphere $A_s=4\pi r^2$

Surface area of cube $A_c=6a^2$

Total emmisive remains same

Thus $P=A\epsilon \sigma T^4$

$A_sT_s^4=A_cT_c^4$

$\frac{T_c^4}{T_s^4}=\frac{A_s}{A_c}$

$\frac{T_c^4}{T_s^4}=\frac{1}{2}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{3}}$

$\frac{T_c}{T_s}=\frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=T_s\times \frac{1}{2^{0.25}}\times \left ( \frac{4\pi}{3}\right )^{\frac{1}{12}}$

$T_c=614\times \frac{1.12679}{1.189}$

$T_c=581.87 K$

3 0

3 years ago

What would happen if an automobile moving at a high speed suddenly comes to a stop? How does this relate to Newton's laws and wh

insens350 [35]

If an automobile moving at high speed suddenly comes to a stop, you would have a large change in momentum. This relates to Newton's second law in the form F = delta p / delta t, where p is momentum (mv).
You could lessen the effect of the sudden stop on the passengers by changing the average force exerted on them. If you look at Newton's second law again, you can see that given some delta p, you can decrease F by increasing delta t. What this means is that if you increase the length of time over which the change in momentum occurs, you can decrease the average force exerted to obtain that change in momentum. This is the reason why landing on a soft cushion is preferable to landing on a concrete surface. The cushion gives way to any object falling on it while still providing some resistance (you don't stop as abruptly), so while your change in momentum is the same in both cases, you have a larger delta t in the case of the cushion.

6 0

3 years ago

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

nata0808 [166]

<u>Answer:</u>

Golf ball will go a maximum of 270.36 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Considering upward vertical motion of projectile.

In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

0 = u sin θ - gt

t = u sin θ/g

Total time for vertical motion is two times time taken for upward vertical motion of projectile.

So total travel time of projectile = 2u sin θ/g

Horizontal motion:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

So range of projectile, $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

Now in the given problem

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

u = 51.5 m/s, for maximum range θ = 45⁰

So maximum distance reached = $\frac{51.5^2sin(2*45)}{9.81}=270.36 meter$

So it will go a maximum of 270.36 meter.

5 0

3 years ago

Does anyone know how to solve this??

Scrat [10]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wtjfavyw

7 0

3 years ago

What would the answer be ?

Marianna [84]

The second one since you’re changing the soil up by adding different fertilisers. This will be you’re independent variable. And you’re dependent variable is your result = the plant height .
Hope this helps :)

4 0

3 years ago