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s2008m [1.1K]
4 years ago
5

To measure the strength of an earthquake, you can use either a _____ scale or _____ scale.

Physics
2 answers:
allsm [11]4 years ago
8 0
To measure the strength of an earthquake, you can use either a Richter scale or Mercalli scale. Richter scale uses the amplitued of the wave and the distance from the source. Mercalli scale uses observations of people and is not considered to be scientific as Richter scale.
Shalnov [3]4 years ago
7 0

Earthquakes are measured with the Richter Magnitude Scale developed by Charles F. Richter of California Institute of Technology in 1935. Here is a summary of the Richter scale:

Magnitude less than 3.5: Generally not felt, but recorded.

Magnitude 3.5-5.4: Often felt, but rarely causes damage.

Magnitude under 6.0: At most slight damage to well-designed buildings over small regions.

Magnitude 6.1-6.9: Can be destructive in areas up to 100 kilometers across where people live.

Magnitude 7.0-7.9: Major earthquake. Can cause serious damage over larger areas.

Magnitude 8 or greater: Great earthquake. Can cause serious damage in areas several hundred kilometers across.

Earthquakes are also measured by the Mercalli Intensity Scale which is a longer, more detailed scale. Both the Richter Scale and the Mercalli Scale can be seen on the U.S. Geological Survey website.



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20. Using the picture, how many neutrons does lithium have?
Igoryamba

Answer:

No. of Neutrons = 3

Explanation:

The atomic number of Lithium is given as 3 in the symbol while the mass number is given as 5.941 which is approximately equal to 6.

Mass Number = No. of Protons + No. of Neutrons = 6

Atomic Number = Number of Electrons = No. of Protons = 3

Therefore,

Mass Number - Atomic Number = (No. of Protons + No. of Neutrons) - No. of Protons

Mass Number - Atomic Number = No. of Neutrons

No. of Neutrons = 6 - 3

<u>No. of Neutrons = 3 </u>

8 0
3 years ago
Which examples are simple machines?
frozen [14]
A hammer and a pulley
4 0
3 years ago
Which factors affect heat transfer between a warm and a cool substance?
andrey2020 [161]

Answer: B

Explanation:

It's not the time it took to heat the substance, so that rules out A and C.

This means that we only have to choose between

B. the area of contact

D. the area of the substances

(since everything else in each of those answers are the same)

Area of contact matters more (e.g. an object with greater surface area is exposed to the air more, will lose/gain heat quicker than an object with less surface area).

5 0
3 years ago
In the picture below, a doctor is having a video conference with a patient. What might be a positive effect of the technology be
GarryVolchara [31]

Answer:

Sorry to say but where  is the photo???

The positive effect of technology being used might be using sethescope or checking BP rate if it is good.

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6 0
3 years ago
A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
Vlad [161]

Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

5 0
3 years ago
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