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Crazy boy [7]
4 years ago
8

Find the mass if the force is 18 N and the acceleration is 2 m/s2

Engineering
2 answers:
Nana76 [90]4 years ago
5 0

Explanation:

f = ma

m = f/a

= 18N / 2m/s2

= 9kg

Rzqust [24]4 years ago
5 0

Answer:

9kg

Explanation:

Step one: write the formula of Force

F= M× A

F- Force

M- Mass - Kg

A- Acceleration - m/s²

Step two: Use the formula

F= M× A

M = F/ A

F= 18N, A= 2m/s²

M= 18N / 2m/s² = 9 kg

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A freeway is being designed for a location in rolling terrain. The expected free-flow speed is 55 mi/h. During the peak hour, it
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3.

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4 years ago
Determine the resolution of a manometer required to measure the velocity of air at 50 m/s using a pitot-static tube and a manome
oksano4ka [1.4K]

Answer:

a)  Δh = 2 cm,  b) Δh = 0.4 cm

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Let's start by using Bernoulli's equation for the Pitot tube, we define two points 1 for the small entry point and point 2 for the larger diameter entry point.

            P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Point 1 is called the stagnation point where the fluid velocity is reduced to zero (v₁ = 0), in general pitot tubes are used  in such a way that the height of point 2 of is the same of point 1

           y₁ = y₂

subtitute

           P₁ = P₂ + ½ ρ v₂²

           P₁ -P₂ = ½ ρ v²

where ρ is the density of fluid  

now we measure the pressure on the included beforehand as a pair of communicating tubes filled with mercury, we set our reference system at the point of the mercury bottom surface

           ΔP =ρ_{Hg} g h - ρ g h

           ΔP =  (ρ_{Hg} - ρ) g h

as the static pressure we can equalize the equations

          ΔP = P₁ - P₂

         (ρ_{Hg} - ρ) g h = ½ ρ v²

         v = \sqrt{\frac{2 (\rho_{Hg} - \rho) g}{\rho } } \ \sqrt{h}

in this expression the densities are constant

        v = A  √h

       A =\sqrt{\frac{2(\rho_{Hg} - \rho ) g}{\rho } }

 

They indicate the density of mercury rhohg = 13600 kg / m³, the density of dry air at 20ºC is rho air = 1.29 kg/m³

we look for the constant

        A = \sqrt{\frac{2( 13600 - 1.29) \ 9.8}{1.29} }

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we substitute

       v = 454.55 √h

to calculate the uncertainty or error of the velocity

         h = \frac{1}{454.55^2} \ v^2

       Δh = \frac{dh}{dv}   Δv

       \frac{\Delta h}{h } = 2 \ \frac{\Delta v}{v}

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a) uncertainty 2.5 m / s ( 0.05)

        \frac{\delta v}{v} = 0.05

       \frac{\Delta h}{h} = 2 0.05  

       Δh = 0.1 h

       Δh = 0.1  20 cm

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        \frac{\Delta h}{h} =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

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