Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^<span> 4) = </span>8.00×10^−3. That is how you get your answer
Answer:

Explanation:
For the simple pendulum problem we need to remember that:
,
where
is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:
,
where
is the angular frequency.
There is also an equation that relates the oscillation period and the angular frequeny:
,
where T is the oscillation period. Now, we can easily solve for L:

<span>1. It must be an object which independently orbits the Sun (this means moons can't be considered planets, since they orbit planets)
2. It must have enough mass that its own gravity pulls it into a spheroidal shape.
3. </span><span>It must be large enough to "dominate" its orbit (i.e. its mass must be much larger than anything else which crosses its orbit).</span>
Answer:
it moves 25 inches.
Explanation:
the east west bit isn't important, ignore it. if an ant starts at 6 then moves to 19 then we need to subtract 19 from 6, that's 13. then it moves to 7. the difference between 19 and 7 is 12. add that to 13 and you get 25. it's important to remember that there is no such thing as negative distance. if it moved, then it counts.
Answer: Option (C) is the correct answer.
Explanation:
As we know that metals are able to conduct electricity so, when a negatively charges rod is kept closer to the left sphere then electrons will enter the sphere.
Since, like charges repel each other. Hence, some of the negative changes from the rod will repel the negative charges of left sphere.
As both left and right spheres are touching each other so, the electrons will move towards the right sphere. As a result, there will be too many electrons (negative charge) present on the right sphere and very less electrons present in the left sphere.
Thus, we can conclude that the statement right sphere is negatively charged, another is charged positively, is true.