Answer:
Part 1) Time of travel equals 61 seconds
Part 2) Maximum speed equals 39.66 m/s.
Explanation:
The final speed of the train when it completes half of it's journey is given by third equation of kinematics as

where
'v' is the final speed
'u' is initial speed
'a' is acceleration of the body
's' is the distance covered
Applying the given values we get

Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as

Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance
Thus total time of journey equals
Part b)
the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals 
Answer:
A) some of the rocks energy is transformed to thermal energy
Explanation:
If we neglect air resistance during the fall of the rock, than the mechanical energy of the rock (which is sum of its potential energy and its kinetic energy) would be constant during the entire motion, so the total energy of the rock at the top would be the same as the sum of its potential energy and kinetic energy at the bottom.
However, this not occurs, due to the presence of air resistance. In fact, air resistance acts against the fall of the rock, and because of the friction between the molecules of air and the surface of the rock, the rock loses part of its energy. This energy is converted into thermal energy of the molecules of the air.
Explanation:
1. Movement of water, food and mineral salts in plants
2. Absorption of water by towels when wiping our bodies
3. It is used to absorb ink using a blotting paper or tissue
I would think that you would have to do 42/2=21Hz, but I'm not sure...
To solve this, we use the Wien's Displacement Law as shown in the attached picture. First, convert the temperature to Kelvin.
C to F:
C = (F - 32)*5/9
C = (325 - 32)*5/9 = 162.78 °C
C to K:
K = C + 273
K = 162.78 + 273 = 435.78 K
λmax = 2898/435.78 =
<em>6</em><em>.65 μm</em>