Answer:

Explanation:
Mass of a hockey puck, m = 0.17 kg
Force exerted by the hockey puck, F' = 35 N
The force of friction, f = 2.7 N
We need to find the acceleration of the hockey puck.
Net force, F=F'-f
F=35-2.7
F=32.3 N
Now, using second law of motion,
F = ma
a is the acceleration of the hockey puck

So, the acceleration of the hockey puck is
.
Answer:
The y-component of the normal force is 45.74 N.
Explanation:
Given that,
Mass of the crate, m = 5 kg
Angle with hill, 
We need to find the y component of the normal force. We know that the y component of the normal force is given by :

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.
Answer:
β2= β1+10*f
Explanation:
comparing β2 and β1, it is said that β2 is increased by a factor of f.
for each factor of f, there is a 10*f dB increase.
therefore if the β1 is increases by an intensity of factor f
the new intensity would be β1+ 10*f