Answer : The correct option is, (D) 278 K
Explanation :
We are given temperature .
Now the conversion factor used for the temperature is,
where, K is kelvin and is Celsius.
Now put the value of temperature, we get
Therefore, the temperature 278 K is equal to the
Answer:
0.1575 m/s^2
Explanation:
Solution:-
- Acceleration ( a ) is expressed as the rate of change of velocity ( v ).
- We are given that the trains starts from rest i.e the initial velocity ( vo ) is equal to 0. Then the train travels from reference point ( so = 0 ) to ( sf = 5.6 km ) from the reference.
- During the travel the train accelerated uniformly to a speed of ( vf =42 m/s ).
- We will employ the use of 3rd kinematic equation of motion valid for constant acceleration ( a ) as follows:
- We will plug in the given parameters in the equation of motion given above:
Answer: the acceleration during the first 5.6 km of travel is 0.1575 m / s^2