Answer:
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Explanation:
We can answer this question by using Kepler's second law of planetary motion, which states that:
"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"
This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).
In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Answer:
So the specific heat of the liquid B is greater than that of A.
Explanation:
Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,

where:
m = mass of the body
c = specific heat of the body
change in temperature of the body
The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.
So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.


So the specific heat of the liquid B is greater than that of A.
Answer:
my name is Deepika Pandey anion I am 9 years old my father name is Dinesh Pandey my name is and my sister name is sister name is a
Path length is 2*pi*0.4=2.512
Speed=distance/time
Speed =2.512/0.2=12.56m/s
Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.