Answer:
0,93 atm
Explanation:
For this we will use PV = nRT
P is what we want to find
V = 1 L
n =
= 0,038 moles
R = 0,082 
T = 25°C = 298,15 K
P * 1 = 0,038 *0,082 * 298,15
P = 0,93 atm
Kinetic energy than parked
An applied force must be applied to move a load. This applied force must be large enough to overcome any opposing forces in order to move the load.
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