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KiRa [710]
3 years ago
9

Which scale would most likely be used to tell how much earthquake damage was done to homes and other buildings? a. the Richter s

cale b. the Mercalli scale c. the moment magnitude scale d. the seismic scale Please select the best answer from the choices provided
Physics
2 answers:
larisa86 [58]3 years ago
8 0

Mercalli scale, Richter scale, and moment magnitude scale are all scales that are used for measuring earthquakes. However each of them measures different parameter. The Mercalli scale would most likely be used to tell how much earthquake damage was done to homes and other buildings. This scale measures by observation along the faults and g ives the rating of the earthquake depending on the damage done in the area.

Colt1911 [192]3 years ago
5 0
B. The Mercalli scale
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Can a body be said to be at rest and in motion at the same time?​
andrew-mc [135]

Explanation:

Rest and motion are relative terms. A body at rest with respect to one observer may be in motion with respect to another observer. For example a person sitting in a moving train is at rest with respect to the train.

5 0
3 years ago
Which are different ways ultrasound technology may be used? Check all that apply.
Kamila [148]

Answer:

Detecting kidney stones and cancer, visualizing fetus in the womb

Explanation:

Ultrasound technology can be used in detecting kidney stones and cancer throat and also visualizing a fetus in the womb. The use of ultrasound for detecting kidney stones started around 1961. Ultrasound has also been used in detecting various types of cancer including throat cancer. The use of ultrasound in knowing the gender of fetus in hospitals has been used for a long time. However, it's not used for cleaning dirty surgical tools.  therefore, choice A, D and E apply.

  • Detecting kidney stones
  • Detecting throat cancer
  • Visualizing a fetus in the womb

8 0
3 years ago
Read 2 more answers
Which best describes why Keplers observation of planetary motion is a law instead of a theory
svet-max [94.6K]

Kepler's first law - sometimes referred to as the law of ellipses - explains that planets are orbiting the sun in a path described as an ellipse. An ellipse can easily be constructed using a pencil, two tacks, a string, a sheet of paper and a piece of cardboard. Tack the sheet of paper to the cardboard using the two tacks. Then tie the string into a loop and wrap the loop around the two tacks. Take your pencil and pull the string until the pencil and two tacks make a triangle (see diagram at the right). Then begin to trace out a path with the pencil, keeping the string wrapped tightly around the tacks. The resulting shape will be an ellipse. An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points is a constant. The two other points (represented here by the tack locations) are known as the foci of the ellipse. The closer together that these points are, the more closely that the ellipse resembles the shape of a circle. In fact, a circle is the special case of an ellipse in which the two foci are at the same location. Kepler's first law is rather simple - all planets orbit the sun in a path that resembles an ellipse, with the sun being located at one of the foci of that ellipse.


5 0
3 years ago
Read 2 more answers
A tuned mass damper for a skyscraper consists of a mass–spring system with spring constant 0.288 M/m. What should be its mass if
Morgarella [4.7K]

Answer:

m = 0.255kg

Explanation:

from the formular of a mass - spring system

T = 2π√m/k

making m as the subject of formular

m = T² k/4π²

T =5.91s

k = 0.288 N/m

m = 10.059/39.489

m = 0.255kg

3 0
4 years ago
You are driving your car, and the traffic light ahead turns red. you apply the brakes for 2.26 s, and the velocity of the car de
Lorico [155]

Let the initial velocity of the car be u.

Final velocity of the car (v) = 5.43 m/s

deceleration (a) = - 2.78 m/s^2

Time taken (t) = 2.26 s

Using the first equation of motion:

v = u + at

u = v - at

u = 5.43 - (-2.78 \times 2.26)

u = 5.43 + 6.28

u = 11.71 m/s

Let the car's displacement be x.

Using second equation of motion:

x = ut + \frac{1}{2}at^2

x = 11.71 \times 2.26 - \frac{1}{2} \times 2.78 \times 2.26^2

x = 26.4646 - 7.0995

x = 19.3651 meters

Hence, the displacement of the car is 19.36 meters

6 0
3 years ago
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