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faltersainse [42]
3 years ago
8

The force, F, of the wind blowing against a building is given by where V is the wind speed, rho the density of the air, A the cr

oss-sectional area of the building, and CD is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.
Physics
1 answer:
Sveta_85 [38]3 years ago
7 0

Answer:

dimensions of the drag coefficient is [M^0 L^0 T^0]

Drag coefficient is a dimensionless quantity

Explanation:

force is given byF=\frac{C_{D}  \rho V^2 A}{2}

we get expression for drag coefficient C_{D} =\frac{2F}{\rho V^2 A}

By substituting the dimensions  of the F,V,A and density , we get

C_{D} =\frac{[F]}{[\rho ][V]^2[A]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L T^{-1}]^2[L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L^2 T^{-2}][L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[MLT^{-2}]}\\C_{D}=[M^0 L^0 T^0]

Drag coefficient is a dimensionless

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Answer: B

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What is know as law of inertia? ​
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Explanation:

  • Law of inertia, also called Newton's first law, postulate in physics that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.
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5 0
3 years ago
How much work must be done to stop a 1100-kg car traveling at 112 km/h?(Hint: You will need to convert the speed first.)Answer:
zimovet [89]

According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:

W=\Delta K

Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:

K=\frac{1}{2}mv^2

Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:

\begin{gathered} K=\frac{1}{2}(1100kg)\left(112\frac{km}{h}\times\frac{1\frac{m}{s}}{3.6\frac{km}{h}}\right)^2=532,345.679...J \\  \\ \therefore K\approx532,346J \end{gathered}

Therefore, the answer is: 532,346 J.

5 0
1 year ago
Plz help it really easy
Mrac [35]

The answer is high to low.

3 0
3 years ago
Read 2 more answers
If you have two uncertainties, and they are from two different sources and contribute to the uncertainty of a measurement, what
Darya [45]

The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.

                           Δm = ∑  | \frac{dm}{dx_i} | \ \Delta x_i

Physical quantities are precise values ​​of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the  cases, all the errors add up.

If m is the calculated quantity, x_i the measured values ​​and Δx_i the uncertainty of each value, the total uncertainty is

                      Δm = ∑  | \frac{dm}{dx_i } | \ \Delta x_i    | dm / dx_i | Dx_i

               

for instance:

If the magnitude is  a average of two magnitudes measured each with a different error

                     m = \frac{m_1+m_2}{2}

                     Δm = | \frac{dm}{dx_1} |  Δx₁ + | \frac{dm}{dx_2} | Δx₂

                     \frac{dm}{dx_1} = ½

                     \frac{dm}{dx_2} = ½

                     Δm = \frac{1}{2} Δx₁ + ½ Δx₂

                     Δm = Δx₁ + Δx₂

In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.

Learn more about propagation errors here:

brainly.com/question/17175455

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