Explanation:
A compound is a pure substance composed of two or more different atoms chemically bonded to one another. A compound can be destroyed by chemical means. It might be broken down into simpler compounds, into its elements or a combination of the two.
Answer:

Explanation:
Given that

We know that acceleration a given as




We know that



So the magnitude of force F

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)
PART A) By definition the frequency in a spring is given by the equation

Where,
m = mass
k = spring constant
Our values are,
k=1700N/m
m=5.3 kg
Replacing,


PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

Where,
k = Spring constant
m = mass
y = Vertical compression
v = Velocity
This expression is equivalent to,

Our values are given as,
k=1700 N/m
V=1.70 m/s
y=0.045m
m=5.3 kg
Replacing we have,

Solving for A,



PART C) Finally, the total mechanical energy is given by the equation



Answer:
Art
Explanation:
Polly's line is linear, while arts line is going up with constant velocity. There for art is going faster.
Answer:
48 m
Explanation:
Two trains traveling towards one another on a straight track are 300m apart when the engineers on both trains become aware of the impending Collision and hit their brakes. The eastbound train, initially moving at 97.0 km/h Slows down at 3.50ms^2. The westbound train, initially moving at 127 km/h slows down at 4.20 m/s^2.
The eastbound train
First convert km/h to m/s
(97 × 1000)/3600
97000/3600
26.944444 m/s
As the train is decelerating, final velocity V = 0 and acceleration a will be negative. Using third equation of motion
V^2 = U^2 - 2as
O = 26.944^2 - 2 × 3.5 S
726 = 7S
S = 726/7
S1 = 103.7 m
The westbound train
Convert km/h to m/s
(127×1000)/3600
127000/3600
35.2778 m/s
Using third equation of motion
V^2 = U^2 - 2as
0 = 35.2778^2 - 2 × 4.2 × S
1244.52 = 8.4S
S = 1244.52/8.4
S2 = 148.2 m
S1 + S2 = 103.7 + 148.2 = 251.86
The distance between them once they stop will be
300 - 251.86 = 48.14 m
Therefore, the distance between them once they stop is 48 metres approximately.