Answer: Option <em>a.</em>
Explanation:
Kepler's 2nd law of planetary motion states:
<em>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</em>
It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.
Velocity includes direction
Meteors are burning up in the mesosphere. The meteors make it through the exosphere and thermosphere without much trouble because those layers don't have much air. But when they hit the mesosphere, there are enough gases to cause friction and create heat.
Weight of the 120kg mass object on the moon
It is a fact that the gravity on the Moon is (1/6)th that on the Earth.
Assuming g ≈ 10 m/s² on the earth.
W = m*(g moon) = 120 * (10/6) = 200N.
Weight on moon = 200N.
The object would weigh approximately 200N on the moon.. The force of gravity acting on the object on the moon is the same as the weight of the object on the moon.
We can't find our answer from your options. Regards
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5CFor%20%5C%20120%20dB%5C%5C%5C%5C120%20%3D%2010Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C12%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C10%5E%7B12%7D%20%3D%20%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5C%5C%5C%5CI%20%3D%2010%5E%7B12%7D%20%5C%20%5Ctimes%20%5C%2010%5E%7B-12%7D%5C%5C%5C%5CI%20%3D%201%20%5C%20W%2Fm%5E2)
![For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2](https://tex.z-dn.net/?f=For%20%5C%2070.7%20dB%5C%5C%5C%5C70.7%20%3D%2010Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C7.07%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5D%5C%5C%5C%5C10%5E%7B7.07%7D%20%3D%20%5Cfrac%7BI%7D%7B1%2A10%5E%7B-12%7D%7D%5C%5C%5C%5CI%20%3D%2010%5E%7B7.07%7D%20%5C%20%5Ctimes%20%5C%2010%5E%7B-12%7D%5C%5C%5C%5CI%20%3D%201%20%5Ctimes%20%5C%2010%5E%7B-4.93%7D%20%5C%20W%2Fm%5E2)
The second distance, r₂, can be determined from sound intensity formula given as;

Therefore, the second distance of the sound from the source is 431.78 m.