Answer:
k = 45.95 N/m
Explanation:
First, we will find the launch speed of the ball by using the formula for the horizontal range of the projectile.

where,
Vo = Launch Speed = ?
R = Horizontal Range = 5.3 m
θ = Launch Angle = 35°
Therefore,

v₀² = 55.33 m²/s²
Now, we know that the kinetic energy gain of ball is equal to the potential energy stored by spring:

where,
k = spring constant = ?
x = compression = 17 cm = 0.17 m
m = mass of ball = 24 g = 0.024 kg
Therefore,

<u>k = 45.95 N/m</u>
If you are talking about sound frequency you need to consider what area you are in because in a concert hall it is big and helps the sound spread but in an airplane it is the opposite.
Operant conditioning, sometimes called <em>instrumental learning</em>, was first extensively studied by Edward L. Thorndike, who observed the behavior of cats trying to escape from home-made puzzle boxes.
Hope this helps!
b. the forces of attraction among them limit their motion.
Answer:
C. 85%
Explanation:
A cylinder fitted with a piston exists in a high-pressure chamber (3 atm) with an initial volume of 1 L. If a sufficient quantity of a hydrocarbon material is combusted inside the cylinder to produce 1 kJ of energy, and if the volume of the chamber then increases to 1.5 L, what percent of the fuel's energy was lost to friction and heat?
A. 15%
B. 30%
C. 85%
D. 100%
work done by the system will be
W=PdV
p=pressure
dV=change in volume
3tam will be changed to N/m^2
3*1.01*10^5
W=3.03*10^5*(1.5-1)
convert 0.5L to m^3
5*10^-4
W=3.03*10^5*5*10^-4
W=152J
therefore
to find the percentage used
152/1000*100
15%
100%-15%
85% uf the fuel's energy was lost to friction and heat