Answer:

Explanation:
R = Horizontal range of projectile = 75 m
v = Velocity of projectile = 37 m/s
g = Acceleration due to gravity = 
Horizontal range is given by

The angle at which the arrow is to be released is
.
Answer:
a) 4.49Hz
b) 0.536kg
c) 2.57s
Explanation:
This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

for some time t you have:
x=0.134m
v=-12.1m/s
a=-107m/s^2
If you divide the first equation and the third equation, you can calculate w:

with this value you can compute the frequency:
a)

b)
the mass of the block is given by the formula:

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

Finally, the amplitude is:

Answer: Water can either increase or decrease the friction between surfaces.
That is because it is impossible to create a law for the behavior of every single different gas, so creating laws for an ideal gas helps us understand the basic nature of gasses which might or might not differ slightly or a lot. By understanding how an ideal gas works, we can understand how a normal gas works.