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Maurinko [17]
3 years ago
13

An earthquake’s epicenter is _____.

Physics
2 answers:
Evgen [1.6K]3 years ago
4 0
The point in which it originates.
Anastasy [175]3 years ago
4 0
Hi, hope you’re having a good day.

An earthquake epicenter is the point on the earth's surface above the hypocenter (or focus), point in the crust where a seismic rupture begins.

Idk, you could search more on the internet but if you need help understanding more, you can just comment it and I’ll try to reply ASAP.


Anyways, thanks for reading, have an amazing day, and stay happy. ❤️❤️
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James gently releases a ball at the top of a slope But does not push the ball. Space the ball rolls down the slope. Which force
NARA [144]
On an incline, the force causing the ball to move downwards would be gravity. Additionally, the component of gravity causing this ball to move downwards would be mgsintheta.

Hope this helps!
7 0
3 years ago
Read 2 more answers
Whats the first state discovered?​
larisa [96]

Answer:

The First State Was Delaware. Delaware was made a state in December 7, 1787. It was made a state a week before Pennsylvania.

Give Brainliest if you please

4 0
3 years ago
What is the acceleration experiance by a car that takes 10s to reach 27m/s from rest
lina2011 [118]

Magnitude of acceleration = (change in speed) / (time for the change).

Change in speed  =  (27 - 0)  =  27 m/s
Time for the change = 10 s

Magnitude of acceleration =  (27 m/s) / (10 s)  =  2.7 m/s²  .
4 0
3 years ago
A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are
igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

T = I\alpha = 0.303*0.6454 = 0.196 Nm

So the force acting on the other end to generate this torque mush be:

F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N

4 0
3 years ago
(b) Show that for certain incident energies there is 100 percent transmission. Suppose that we model an atom as a one-dimensiona
bixtya [17]

Answer:

Explanation:

100 persent transmission implies that the T=1

Therefore using the previous result we have

1+\frac{\sin^2\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a}{4\frac{E}{V_0}\frac{(E+V_0)}{V_0}}=1


\sin\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a=0\Rightarrow \sqrt{\frac{2m}{\hbar^2}(E+V_0)}=0\Rightarrow E=-V_0


The depth of the well for 100% transmission should be

V_0=-0.7~{\rm{eV}}

7 0
2 years ago
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