Answer:
9.8 m/s/s
Explanation:
The numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall is 9.8 m/s/s. This is called the acceleration due to gravity.
Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
where m is the slope
Comparing equation (1) and (2)

a = m
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s


The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
![[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%2A%20%2822%20-%2011.78%29%20%2A%20%287.30%29%5D%20%20%2B%20%5B%2811.78%20-%200%29%20%2A%20%287.30%29%5D)
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m
Answer:
a)
, b) 
Explanation:
a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

The final velocity is:

b) The change in the kinetic energy of the 13.5 g coin is:
![\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%2813.5%5Ctimes%2010%5E%7B-3%7D%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%2811.9%5Ctimes%2010%5E%7B-2%7D%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D-%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%5Cright%5D)
