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:
Flow Rate = 80 m^3 /hours (Rounded to the nearest whole number)
Explanation:
Given
- Hf = head loss
- f = friction factor
- L = Length of the pipe = 360 m
- V = Flow velocity, m/s
- D = Pipe diameter = 0.12 m
- g = Gravitational acceleration, m/s^2
- Re = Reynolds's Number
- rho = Density =998 kg/m^3
- μ = Viscosity = 0.001 kg/m-s
- Z = Elevation Difference = 60 m
Calculations
Moody friction loss in the pipe = Hf = (f*L*V^2)/(2*D*g)
The energy equation for this system will be,
Hp = Z + Hf
The other three equations to solve the above equations are:
Re = (rho*V*D)/ μ
Flow Rate, Q = V*(pi/4)*D^2
Power = 15000 W = rho*g*Q*Hp
1/f^0.5 = 2*log ((Re*f^0.5)/2.51)
We can iterate the 5 equations to find f and solve them to find the values of:
Re = 235000
f = 0.015
V = 1.97 m/s
And use them to find the flow rate,
Q = V*(pi/4)*D^2
Q = (1.97)*(pi/4)*(0.12)^2 = 0.022 m^3/s = 80 m^3 /hours
Answer:
a) 
b) 
c) 
Explanation:
From the exercise we know that the ball strikes the building 16m away and its final height is 8m more than the initial
Being said that, we can calculate the initial velocity of the ball
a) First we analyze its horizontal motion


(1)
That would be our first equation
Now, we need to analyze its vertical motion


Knowing
in our first equation (1)


Solving for t

So, the ball takes to seconds to get to the other building. Now we can calculate its <u>initial velocity</u>

b) To find the <u>magnitude of the ball just before it strikes the building</u> we need to calculate its x and y components


So, the magnitude of the velocity is:

c) The <u><em>direction of the ball</em></u> is:

Answer:
A = 1.4 m/s²
B = -0.10493 m/s³
a = 1.29507 m/s²
T = 28095.8271 N
T = 1.13198 W
Explanation:
t = Time taken
g = Acceleration due to gravity = 9.81 m/s²
The equation

Differentiating with respect to time

At t = 0

Hence, A = 1.4 m/s²

B = -0.10493 m/s³
At t = 5 seconds

a = 1.29507 m/s²

T = 28095.8271 N
Weight of rocket


T = 1.13198 W