This is a defective, misleading question, and should never be asked in a Physics class.
There is no such thing as the force due to the impact.
If you know how long it takes the clam to stop once it begins to hit the dirt,
then you can calculate the impulse transferred to it, and tease a force out
of that. But the question doesn't give us the time.
It depends on the material of the surface. Was the clam dropped onto dirt ?
Into a dumpster ? Onto grass ? Concrete ? Styrofoam ? Mud ? The answer
is different in each case, and we still need to know the short length of time
AFTER it first encountered whatever surface brought it to rest.
I would kick this question back to the Physics teacher. It's meaningless,
and the longer you try to work on it, the more nonsense you'll plant into
your head that'll need to be dug out later.
The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.
Answer:
1. The nucleus is about 1/2 the size of the atom
Explanation:
Alpha particles are positive charge particles and they are bounced back by the nucleus because nucleus is also same size
Now in present experiment Rutherford found that very few alpha particles are bounced back along same path which shows that very small region inside the nucleus is having positive charge and rest part of the atom is empty.
Now if we found that half of the alpha particles are bounced back then it shows that size of the nucleus is very large now as compare to previous one because only nucleus can bounce back the alpha particles
so correct answer will be
1. The nucleus is about 1/2 the size of the atom
Answer:
I think it's D. open series circuit .
Explanation:
<em>hope</em><em> it</em><em> helps</em><em> you</em><em>!</em>
Answer:
96%
Explanation
Let A the total area of the galaxy, is modeled as a disc:
A = πR^2 = π (25 kpc)^2
And let a be the area that astronomers are able to see:
a = πr^2 = π(5 kpc)^2
The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:
P = 100 a/A = (5/25)^2 = 100/25 = 4%
Therefore, the percentage of the galaxy not included, i.e. not seen is:
(100-4)% = 96%
