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Tamiku [17]
3 years ago
13

A student used a yardstick to model the displacement of rock at a fault during an earthquake. The student bent the yardstick wit

hout breaking it and then let it go. What process did the student show?
A. A moment magnitude vibration
B. Elastic rebound
C. Isostasy
D. The formation of a new fault
Physics
2 answers:
jonny [76]3 years ago
7 0
I think the correct answer would be B. The process of elastic rebound is being shown by the student. It is a theory that is used to explain earthquakes. It focuses on how energy is being spread in times of earthquakes. As the rocks on the fault experiences shift and force, these rocks would be accumulating energy causing it to deform reaching the internal strength and eventually exceeding it. At that moment, a rapid motion would happen along the fault, which releases the energy, then the rocks would go back to its original shape or the undeformed state. This theory is the first theory that sufficiently was able to explain earthquakes.
dlinn [17]3 years ago
6 0

Answer:

For APEX: Elastic Rebound

Explanation:

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Consider two uniform solid spheres where both have the same diameter, but one has twice the mass of the other. how much larger i
egoroff_w [7]
<span>The moment of inertia of the large sphere will be twice that of the smaller sphere.
   The formula for the moment of inertia for a solid sphere is:
 I = (2/5)mr^2
 where
 I = moment of inertia
 m = mass
  r = radius

   Since both spheres have the same diameter, they also have the same radius, so the only change is their mass. And the moment of inertia is directly proportional to their mass as shown by the above formula. So the sphere with twice the mass will have twice the moment of inertia, or 2 times.</span>
5 0
3 years ago
An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
How much heat is needed to raise the temperature of 50.0 g of water by 25.0°C
love history [14]

Answer:

Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

c

=

4.18

J

g

∘

C

Now, let's assume that you don't know the equation that allows you to plug in your values and find how much heat would be needed to heat that much water by that many degrees Celsius.

Take a look at the specific heat of water. As you know, a substance's specific heat tells you how much heat is needed in order to increase the temperature of

1 g

of that substance by

1

∘

C

.

In water's case, you need to provide

4.18 J

of heat per gram of water to increase its temperature by

1

∘

C

.

What if you wanted to increase the temperature of

1 g

of water by

2

∘

C

? You'd need to provide it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

=

increase by 2

∘

C



2

×

4.18 J

To increase the temperature of

1 g

of water by

n

∘

C

, you'd need to supply it with

increase by 1

∘

C



4.18 J

+

increase by 1

∘

C



4.18 J

+

...

=

increase by n

∘

C



n

×

4.18 J

Now let's say that you wanted to cause a

1

∘

C

increase in a

2-g

sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

To cause a

1

∘

C

increase in the temperature of

m

grams of water, you'd need to supply it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

+

,,,

=

for m g of water



m

×

4.18 J

This means that in order to increase the temperature of

m

grams of water by

n

∘

C

, you need to provide it with

heat

=

m

×

n

×

specific heat

This will account for increasing the temperature of the first gram of the sample by

n

∘

C

, of the the second gram by

n

∘

C

, of the third gram by

n

∘

C

, and so on until you reach

m

grams of water.

And there you have it. The equation that describes all this will thus be

q

=

m

⋅

c

⋅

Δ

T

, where

q

- heat absorbed

m

- the mass of the sample

c

- the specific heat of the substance

Δ

T

- the change in temperature, defined as final temperature minus initial temperature

In your case, you will have

q

=

100.0

g

⋅

4.18

J

g

∘

C

⋅

(

50.0

−

25.0

)

∘

C

q

=

10,450 J

Rounded to three sig figs and expressed in kilojoules, t

Explanation:

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3 years ago
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Jobisdone [24]
Let the observer be 'd' distance away from the thunderstorm and let light take 't' time to reach the observer
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For light,
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For sound,
           339 = d/(t + 30)       -2

Putting value from 1 in 2.
               d = 10^4 m(approx)
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