Answer:
The drag coefficient is
Explanation:
From the question we are told that
The density of air is
The diameter of bottom part is
The power trend-line equation is mathematically represented as
let assume that the velocity is 20 m/s
Then
The drag coefficient is mathematically represented as
Where
is the drag force
is the density of the fluid
is the flow velocity
A is the area which mathematically evaluated as
substituting values
Then
Answer:
meters
Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function
The maxima of the function occurs when the slope is zero. i.e.
Hence the maxima occurs at t=1.63 seconds
The maximum value of f is
hence maximum height is found to be
meters
Answer:
Temperature at the exit =
Explanation:
For the steady energy flow through a control volume, the power output is given as
Inlet area of the turbine =
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume
as
for Ideal gasses, the enthalpy change can be calculated using the formula
hence we have
<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>
evaluating the above equation, we have
Hence, the temperature at the exit =
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m
K.E = 1/2 x 0.17 x
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/
To calculate the final velocity, let us use third equation of motion.
= + 2as
= + 2 x 0.3 x 61
= 36 + 36
= 72
V =
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
Learn more about dynamics here: brainly.com/question/402617