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Lemur [1.5K]
3 years ago
9

Do you think seismographs predict earthquakes or measure earthquakes, explain your answer?

Physics
2 answers:
Delvig [45]3 years ago
6 0
An earthquake prediction must define 3 elements: 1) the date and time, 2) the location, and 3) the magnitude.

Yes, some people say they can predict earthquakes, but here are the reasons why their statements are false:

They are not based on scientific evidence, and earthquakes are part of a scientific process. For example, earthquakes have nothing to do with clouds, bodily aches and pains, or slugs.
They do not define all three of the elements required for a prediction.
Their predictions are so general that there will always be an earthquake that fits; such as, (a) There will be a M4 earthquake somewhere in the U.S. in the next 30 days. (b) There will be a M2 earthquake on the west coast of the U.S. today.
Yuki888 [10]3 years ago
5 0

Answer:

No. Neither the USGS nor any other scientists have ever predicted a major earthquake. We do not know how, and we do not expect to know how any time in the foreseeable future.

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Find the speed of a rock being thrown 20.5 meters (m) to the left in 4.0s.
erik [133]

Answer:

5.125

Explanation:

formula for speed is distance/time distance measured in metres

8 0
1 year ago
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Which is not one of the three forms of energy that travels to earth
Tju [1.3M]
Is there options for this??
5 0
2 years ago
How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?
SVEN [57.7K]

Time = (distance) / (speed)

<em></em>

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

8 0
3 years ago
A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.
Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

 

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

      -10 + 12 sin 60 + Fₓ = 0

        Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

       12 cos 60 - 6 + F_y = 0

        F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

  • Form of coordinates F = -0.39 i ^ N
  • Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

       F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}  

       F = 0.39N

We use trigonometry for the angle.

       tan \theta = \frac{F_y}{F_x}

       tan θ=  0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

        θ = 180 + θ'

        θ = 180 + 0

        θ = 180º

C) if the three forces can be moved and the maximum force occurs when they are all linear.

          10+ 6 + 6 + F = 0

          F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

          a = \frac{F}{m}  

          a = \frac{24}{6}  

         a =  4 m / s²

The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

        v = v₀ - a t

Final velocity when stopped is zero

         t = \frac{0-v_o}{a}

         t = 100/4

         t = 25 s

In conclusion using Newton's second law and the kinematics relations we can find the results for the questions about the forces and the motion of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

Learn more about Newton's second law here: brainly.com/question/25545050

3 0
2 years ago
A 40kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hill side?
bagirrra123 [75]

Answer:

<h2>39.2 m</h2>

Explanation:

The height of the hill side can be found by using the formula

h =  \frac{p}{m}  \\

p is the potential energy

m is the mass

From the question we have

h =  \frac{1568}{40}  =  \frac{196}{5}  \\

We have the final answer as

<h3>39.2 m</h3>

Hope this helps you

5 0
2 years ago
Read 2 more answers
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