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kykrilka [37]
2 years ago
10

about how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?

Physics
1 answer:
OverLord2011 [107]2 years ago
8 0

Answer:

For example, an earthquake of magnitude 5.5 releases about 32 times as much energy as an earthquake measuring 4.5. Another way to look at this is that it takes about 900 magnitude 4.5 earthquakes to equal the energy released in a single 6.5 earthquake.

Explanation:

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These brain chemicals, including adrenaline, dopamine and cortisol, stimulate a natural high and give you a boost of energy that makes you feel more alert, alive and able to scream your lungs out

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Metric conversions
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1:-

  • 1L=1000mL

\\ \sf\longmapsto 45.23L

\\ \sf\longmapsto 45.23\times 1000

\\ \sf\longmapsto 45230mL

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\\ \sf\longmapsto .035hL

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\\ \sf\longmapsto 27.32

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3 years ago
A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav
oksano4ka [1.4K]

Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

t = 0.99 s

Now, we can calculate the initial speed:

8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

<u>v0 = 10 m/s</u>

The takeoff speed is 10 m/s

b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

<u>y = 1.2 m</u>

The maximum height above the ground is 1.2 m.

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3 years ago
Scientists can use the location and depth of where a fossil is found to estimate the age of the fossil. To do this, they use
Roman55 [17]

Answer:

C the Law of Superposition

Explanation:

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When we find fossils at a particular depth we can relatively date a rock based on the strata we find them. A fossil in place in bottom beds will be older than the one in the topmost layer. Fossils also succeed one another in a definite pattern according to the principle of fossil and fauna succession.

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