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Step2247 [10]
3 years ago
14

Before the widespread use of computers, how was the epicenter of an earthquake determined?

Physics
2 answers:
Nezavi [6.7K]3 years ago
7 0
 Data was used from three different stations. Where their data overlapped was where the epicenter was located.
it is called triangulation 
STALIN [3.7K]3 years ago
7 0

Data was used from three different stations. Where their data overlapped was where the epicenter was located.


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a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the
irinina [24]
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


7 0
3 years ago
An object is accelerating at a constant rate due to a force being applied.what can be done to the force to decrease the objects
Sliva [168]

Answer:

y

Explanation:

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3 years ago
What factor might be contributing to climate change
kirza4 [7]

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6 0
4 years ago
si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p
forsale [732]

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

E_{kin}=E_{pot}

Donde:

E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h

Ahora reemplazando:

\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} }   \\\\v=5.24[m/s]

6 0
3 years ago
You have an initial velocity of -3.0 m/s. You then experience an acceleration of 2.5 m/s2 for 9.0s; what
Tatiana [17]
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3 years ago
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