Answer:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;

(ii) The linear speed of the small metal object is calculated as;

law of conservation of energy
aka the first law of thermodynamics
Answer: c. the molecules with the highest energy evaporate first, lowering the temperature of the sample
Explanation:
The process by which liquid starts to change into vapor phase at any temperature is known as evaporation.
During evaporation , the molecules which possess higher energies escape from the upper layer into vapor phase. the molecules which escape draw energy from surroundings and thus decrease the energy of the surroundings and hence lead to decrease in temperature.
As temperature of the system is directly proportional to the energy of the system , thus decrease in energy leads to decrease in temperature.

K.E. = Kinetic energy
T = temperature
R= gas constant
Answer: A.AB
Explanation:
This Velocity vs Time graph shows the acceleration of a body or object, since acceleration is the variation of velocity in time.
As we can see in the attached image, the graph can be divided in four segments:
OA: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a positive slope, hence we are dealing with a positive uniform acceleration.
AB: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be positive. This means the <u>acceleration is nonuniform and positive.</u>
BC: In this segment the acceleration is changing at a nonuniform rate, since in this part it is not possible to calculate the slope. However if this were uniform, the slope woul be negative. This means the acceleration is nonuniform and negative.
CD: In this segment the acceleration is changing at a uniform rate. In addition we can see it has a negative slope, hence we are dealing with a negative uniform acceleration.
From all these segments, the only one that fulfils the nonuniform positive acceleration condition is option A:
Segment AB