Answer: downward velocity = 6.9×10^-4 cm/s
Explanation: Given that the
Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m
Where radius r = 2.5 × 10^-5 m
Density = 1200 kg/m^3
Area of a sphere = 4πr^2
A = 4 × π× (2.5 × 10^-5)^2
A = 7.8 × 10^-9 m^2
Volume V = 4/3πr^3
V = 4/3 × π × (2.5 × 10^-5)^3
V = 6.5 × 10^-14 m^3
Since density = mass/ volume
Make mass the subject of formula
Mass = density × volume
Mass = 1200 × 6.5 × 10^-14
Mass M = 7.9 × 10^-11 kg
Using the formula
V = sqrt( 2Mg/ pCA)
Where
g = 9.81 m/s^2
M = mass = 7.9 × 10^-11 kg
p = density = 1200 kg/m3
C = drag coefficient = 24
A = area = 7.8 × 10^-9m^2
V = terminal velocity
Substitute all the parameters into the formula
V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]
V = sqrt[ 1.54 × 10^-9/2.25×10-4]
V = 6.9×10^-6 m/s
V = 6.9 × 10^-4 cm/s
Answer:
When a pilot pushes the top of the right pedal, it activates the brakes on the right main wheel/wheels, and when the pilot pushes the top of the left rudder pedal, it activates the brake on the left main wheel/wheels. The brakes work in a rather simple way: they convert the kinetic energy of motion into heat energy.
Answer:
transmission bandwidth required is very large.
Explanation:
Answer:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12
Explanation:
A cubic has the form:
y = ax³ + bx² + cx + d
Given four points, we can write a system of equations:
1 = a + b + c + d
1/2 = 8a + 4b + 2c + d
1/3 = 27a + 9b + 3c + d
1/4 = 64a + 16b + 4c + d
Solving this algebraically would be time-consuming, but we can use matrices to make it easy.
![\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right]\left[\begin{array}{cccc}a\\b\\c\\d\end{array}\right]=\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C8%264%262%261%5C%5C27%269%263%261%5C%5C64%2616%264%261%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Da%5C%5Cb%5C%5Cc%5C%5Cd%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%5C%5C1%2F2%5C%5C1%2F3%5C%5C1%2F4%5Cend%7Barray%7D%5Cright%5D)
First, we find the inverse of the coefficient matrix. This is messy to do by hand, so let's use a calculator:
![\left[\begin{array}{cccc}1&1&1&1\\8&4&2&1\\27&9&3&1\\64&16&4&1\end{array}\right] ^{-1} =-\frac{1}{12}\left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%261%261%5C%5C8%264%262%261%5C%5C27%269%263%261%5C%5C64%2616%264%261%5Cend%7Barray%7D%5Cright%5D%20%5E%7B-1%7D%20%3D-%5Cfrac%7B1%7D%7B12%7D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D2%26-6%266%26-2%5C%5C-18%2648%26-42%2612%5C%5C52%26-114%2684%26-22%5C%5C-48%2672%26-48%2612%5Cend%7Barray%7D%5Cright%5D)
Now we multiply by the solution matrix (again using a calculator):
![-\frac{1}{12} \left[\begin{array}{cccc}2&-6&6&-2\\-18&48&-42&12\\52&-114&84&-22\\-48&72&-48&12\end{array}\right]\left[\begin{array}{cccc}1\\1/2\\1/3\\1/4\end{array}\right] =\left[\begin{array}{cccc}-1/24\\5/12\\-35/24\\25/12\end{array}\right]](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B12%7D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D2%26-6%266%26-2%5C%5C-18%2648%26-42%2612%5C%5C52%26-114%2684%26-22%5C%5C-48%2672%26-48%2612%5Cend%7Barray%7D%5Cright%5D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%5C%5C1%2F2%5C%5C1%2F3%5C%5C1%2F4%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D-1%2F24%5C%5C5%2F12%5C%5C-35%2F24%5C%5C25%2F12%5Cend%7Barray%7D%5Cright%5D)
So the cubic is:
y = -1/24 x³ + 5/12 x² − 35/24 x + 25/12
