Answer:
option (D) is correct.
Explanation:
According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.
the kinetic energy of a body is directly proportional to the square of the speed of the body.
As the kinetic energy change, the speed of the body also change.
Option (D) is correct.
Answer:
C.Supersaturated
Explanation:
There are three types of solution:
<u>SATURATED SOLUTION</u>:
It is the solution that contains maximum amount of solute dissolved in a solution in normal conditions.
<u>UNSATURATED SOLUTION</u>:
It is the solution that contains less than the maximum amount of solute dissolved in a solution in normal conditions. It has space for more solute to be dissolved in it.
<u>SUPERSATURATED SOLUTION:</u>
It contains more than the maximum amount of solute dissolved in it. Such a solution has no capacity to dissolve any more solute under any conditions.
Since the sugar is no more dissolving in the tea and has settled down. Therefore, the solution is:
<u>C.Supersaturated</u>
Answer:
Explanation:
a )
Reaction force of the ground
R = mg
= 160 N
Maximum friction force possible
= μ x R
= μ x 160
= .4 x 160
= 64 N .
b )
160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,
Taking moment about top point of ladder
160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3
240 + 444 + 4f = 2700
f = 504 N
c )
Let x be the required distance.
Taking moment about top point of ladder
160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )
240 + 444 x + 1440 = 2700
x = 2.3 m
so man can go upto 2.3 at which maximum friction acts .
Answer:
Explanation:
Learn vocabulary, terms, and more with flashcards, games, and other study tools. ... A jogger sprints 100 m in 13 seconds. What is her average speed? 7.7 m/s ... Kathryn swam 5 complete laps of a 50 m pool. ... stands still for 4 seconds, then continues to walk for 8 meters moving away from the starting point in 6 seconds.
Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.