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Ilya [14]
3 years ago
11

Explain how do you determine the intensity of earthquake​

Physics
2 answers:
Ivanshal [37]3 years ago
6 0

Answer:

The Richter scale measures the largest wiggle (amplitude) on the recording, but other magnitude scales measure different parts of the earthquake. The USGS currently reports earthquake magnitudes using the Moment Magnitude scale, though many other magnitudes are calculated for research and comparison purposes.

Maurinko [17]3 years ago
4 0

Answer:

To measure the magnitude of an earthquake, the American scientist Charles Richter developed a scale in 1935. Known as the <u>Richter scale</u>, it assigns a number based on the height of the waves on a seismogram (the visual output of a seismograph). Seismographs measure ground motion, including the energy released by an earthquake.

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A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.
snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is  v =  3.79 *10^{5} \ m/s  

Explanation:

From the question we are told that

    The  magnitude of the electric field is  E  =  144 \ kV /m   =  144*10^{3} \  V/m

     The magnetic field is  B  =  0.38 \ T

   

The force due to the electric field is mathematically represented as

      F_e =  E  * q

and

The force due to the magnetic field is mathematically represented as

    F_b  =  q * v  *  B * sin(\theta )

Now given that it is perpendicular ,  \theta  = 90

=>   F_b  =  q * v  *  B * sin(90)

=>   F_b  =  q * v  *  B

Now  given that it is not deflected it means that

        F_ e  =  F_b

=>    q *  E = q *  v  * B

=>   v =  \frac{E}{B }

 substituting values

     v =  \frac{ 144 *10^{3}}{0.38 }

     v =  3.79 *10^{5} \ m/s

7 0
3 years ago
A tugboat pulls a ship with a constant net horizontal force of 4.71 × 103 N and causes the ship to move through a harbor. How mu
Studentka2010 [4]

Answer:

21.53 x 10^{6} N

Explanation:

force (F) = 4.71 x 10^{3} N

distance (s) = 4.57 km = 4570 m

how much work is done.

work = force x distance

work = 4.71 x 10^{3} x 4570 = 21.53 x 10^{6} N

8 0
3 years ago
Which sentence correctly describes a friction force? A. It acts in the same direction as the motion of an object. B. It acts in
KatRina [158]

Answer:

B

Explanation:

Friction acts in a direction opposite to the motion of an object

5 0
3 years ago
How can a small human retina detect objects larger than itself?
aliya0001 [1]

Exactly the same way that you can photograph a mountain or a skyscraper
with the itty bitty camera in your smartphone.

Lenses are used to form a tiny image of a gigantic object.


7 0
2 years ago
Read 2 more answers
Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second
vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

I' = \frac{I_0}{2} cos^2 \theta

Where

\theta =Angle From vertical of the axis of the polarizing filter

I_0 = Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

I = \frac{I_0}{2}

Replacing we have that

I = \frac{370}{2}

I = 185W/m^2

Re-arrange the equation,

I'= \frac{I_0}{2}cos^2\theta

Re-arrange to find \theta

cos^2\theta = \frac{2I'}{I_0}

cos^2\theta = \frac{2*138}{370}

\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})

\theta = 0.5282rad

\theta = 30.27\°

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0
3 years ago
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