The answer that is being described above is the ASTEROIDS. The one that we see floating between Mars and Jupiter is what we call the Asteroid Belt. The asteroid belt comprises of different rocky bodies and they also orbit within the solar system. Hope this helps.
Answer:
a) 2.87 m/s
b) 3.23 m/s
Explanation:
The avergare velocity can be found dividing the length traveled d by the total time t.
a)
For the first part we easily know the total traveled length which is:
d = 50.2 m + 50.2 m = 100.4 m
The time can be found dividing the distance by the velocity:
t1 = 50.2 m / 2.21 m/s = 22.7149 s
t2 = 50.2 m / 4.11 m/s = 12.2141 s
t = t1 +t2 = 34.9290 s
Therefore, the average velocity is:
v = d/t =2.87 m/s
b)
Here we can easily know the total time:
t = 1 min + 1.16 min = 129.6 s
Now the distance wil be found multiplying each velocity by the time it has travelled:
d1 = 2.21 m/s * 60 s = 132.6 m
d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m
d = 418.656 m
Therefore, the average velocity is:
v = d/t =3.23 m/s
Answer:
Explanation:
The unknown charge can not remain in between the charge given because force on the middle charge will act in the same direction due to both the remaining charges.
So the unknown charge is somewhere on negative side of x axis . Its charge will be negative . Let it be - Q and let it be at distance - x on x axis.
force on it due to rest of the charges will be equal and opposite so
k3q Q / x² =k 8q Q / (L+x)²
8x² = 3 (L+x)²
2√2 x = √3 (L+x)
2√2 x - √3 x = √3 L
x(2√2 - √3 ) = √3 L
x = √3 L / (2√2 - √3 )
Let us consider the balancing force on 3q
force on it due to -Q and -8q will be equal
kQ . 3q / x² = k3q 8q / L²
Q = 8q (x² / L²)
so charge required = - 8q (x² / L²)
and its distance from x on negative x side = √3 L / (2√2 - √3 )
Answer:
As you know, the denser objects have more weight per unit of volume, this will mean that the force that pulls down these objects is a bit larger.
This will mean that the denser objects will always go to the bottom.
This clearly implies that the red liquid, the one with one of the smaller densities, can not be at the bottom.
There are some cases where a liquid with a small density may become a lot denser as the temperature or pressure changes, and in a case like that, we could see the red liquid at the bottom, but for this case, there is no mention of changes in the temperature nor in the pressure, so this can be discarded.
The only thing that makes sense is that the red part at the bottom is the base of the tube, and has nothing to do with the red liquid.
