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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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2 years ago
Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,
klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble=2\hat{i}-3\hat{k} N

Position vector=r=0.5\hat{j}-2\hat{k} m

a.We have to find the resulting torque on the pebble about origin.

Torque=r\times F

Substitute the values then we get

Torque= (0.5j-2k)\times (2i-3k)

Torque=-k-1.5i-4jN-m

By using i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0

b.r=2i-3k

r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k

Torque about point (2,0,-3)

\tau=(-2i+0.5j+k)\times (2i-3k)

\tau=-6j-k-1.5i+2j=-1.5i-4j-kN-m

6 0
3 years ago
g A bowling ball with a mass of 3.86 kg and a radius of 0.161 m starts from rest at a height of 2.5 m and rolls down a 48.4 o sl
Fynjy0 [20]

Answer:

v=1.5m/s

Explanation:

The gravitational potential energy gets transformed into translational and rotational kinetic energy, so we can write mgh=\frac{mv^2}{2}+\frac{I\omega^2}{2}. Since v=r\omega (the ball rolls without slipping) and for a solid sphere I=\frac{2mr^2}{5}, we have:

mgh=\frac{mv^2}{2}+\frac{2mr^2\omega^2}{2*5}=\frac{mv^2}{2}+\frac{mv^2}{5}=\frac{7mv^2}{10}

So our translational speed will be:

v=\sqrt{\frac{10gh}{7}}=\sqrt{\frac{10(9.8m/s^2)(0.161m)}{7}}=1.5m/s

6 0
2 years ago
Why is velocity and not just speed important to these pilots
Alborosie
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3 years ago
A newspaper turns yellow in sunlight. Is this a physical or chemical change?
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