Answer:1. Roche limit
2.hydrogen
3.atmosphere
4.mercury
5.venus
6.when an object passes the Roche limit, the strength of gravity on the object increases. If the density of the planet is higher, then the object can break up farther away from the planet. If the density is lower, then the Roche limit is located closer to the planet
7.Farther our in the solar system, beyond the frost line, hydrogen was at a low enough temperature that it could condense. This allowed hydrogen to accumulate under gravity, eventually forming the Jovian planets
Explanation:
Answer:
E=0
Explanation:
Electric field due to each thin sheet of charge=\sigma/2\varepsilon
let us say the right plate has positive charge density \varepsilonand left sheet has a negative charge density -\varepsilon .
In the region between the plates,the electric field due to each plate is in same direction,
E=\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=\sigma/\varepsilon
in the region outside the plates, the field due to the plates is in opposite directions
E=-\sigma/2\varepsilon-(-\sigma/2\varepsilon)
E=-\sigma/2\varepsilon+\sigma/2\varepsilon
E=0
Answer:
height is 69.68 m
Explanation:
given data
before it hits the ground = 46 % of entire distance
to find out
the height
solution
we know here acceleration and displacement that is
d = (0.5)gt² ..............1
here d is distance and g is acceleration and t is time
so when object falling it will be
h = 4.9 t² ....................2
and in 1st part of question
we have (100% - 46% ) = 54 %
so falling objects will be there
0.54 h = 4.9 (t-1)² ...................3
so
now we have 2 equation with unknown
we equate both equation
1st equation already solve for h
substitute h in the second equation and find t
0.54 × 4.9 t² = 4.9 (t-1)²
t = 0.576 s and 3.771 s
we use here 3.771 s because 0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls
so take t = 3.771 s
then h from equation 2
h = 4.9 t²
h = 4.9 (3.771)²
h = 69.68 m
so height is 69.68 m
Answer:
Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.
