Answer:
Mass remains constant but weight reduces
Explanation:
Mass is the amount of matter in an object so whether on moon or any other planet, it does not change despite the changes in acceleration.
Weight is a product of mass and acceleration due to gravity, expressed as W=mg where m is the mass, W is weight and g is acceleration. From the above formula, it is evident that when you decrease g, then W also decreases while m is constant. Similarly, when m is constant and g is increased then W also increases.
Therefore, for this case, since g decreases, the weight decreases but mass remains constant.
E = mc^2
m = e/c^2
m = 2.7*10^16/(300000^2)
m = 300000
Answer:
See below
Explanation:
Vertical position is given by
df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)
do =original position = 2 m
vo = original <u>VERTICAL</u> velocity = 0
a = acceleration of gravity = 9.81 m/s^2
THIS BECOMES
0 = 2 + 0 * t - 1/2 ( 9.81)t^2
to show t =<u> .639 seconds to hit the ground </u>
During this .639 seconds it flies horizontally at 10 m/s for a distance of
10 m/s * .639 s =<u> 6.39 m </u>
The FREQUENCY of light remains unchanged once it leaves the source.
Answer:
Yes, it is reasonable to neglect it.
Explanation:
Hello,
In this case, a single molecule of oxygen weights 32 g (diatomic oxygen) thus, the mass of kilograms is (consider Avogadro's number):
After that, we compute the potential energy 1.00 m above the reference point:
Then, we compute the average kinetic energy at the specified temperature:
Whereas 
stands for the Avogadro's number for which we have:
In such a way, since the average kinetic energy energy is about 12000 times higher than the potential energy, it turns out reasonable to neglect the potential energy.
Regards.
