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Paul [167]
3 years ago
8

A particle is moving along a circular path having a radius of 6 in. such that its position as a function of time is given by θ=c

os2t, where θ is in radians and t is in seconds.
Determine the magnitude of the acceleration of the particle when θ= 35 ∘
Engineering
1 answer:
Vera_Pavlovna [14]3 years ago
6 0

Answer:

The angular acceleration is -2.44 rad/s², while the linear acceleration is -14.66 in/s².

Explanation:

First we need to find the time, at the given position. W e are given the position of particle to be:

θ = 35°

Converting it to radians because, the given equation is in radians:

θ = (35°) (π radians/180°)

θ = 0.611 radians

Now, we have the equation:

θ = Cos(2t)

2t = Arc Cos (θ)

2t = Arc Cos (0.611 radians)

t = 0.91/2

t = 0.457 sec

Now, to determine angular acceleration of the particle, we must derivate the equation twice with respect to 't'

Angular Velocity = ω = dθ/dt = -2Sin(2t)

Angular Acceleration = α = -4Cos(2t)

Now, we use the value of t:

α = -4Cos(2 x 0.457)

<u>α = -2.44 rad/s² </u>(negative sign shows decceleration)

Now for linear acceleration, we know that:

a = rα

a = (6 in)(-2.44 rad/s²)

<u>a = -14.66 in/s²</u> (negative sign shows decceleration)

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Calculate the areas under the stress-strain curve (toughness) for the materials shown in Fig. below, (a) plot them as a
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Answer:

Explanation:

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5 0
3 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

Explanation:

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} }

        A = 454.55

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}

Suppose we have a height reading of h = 20 cm = 0.20 m

             

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

       Δh = 2 cm

b) uncertainty 0.5 m / s ( Δv/v= 0.01)

        \frac{\Delta h}{h} =  2 0.01

        Δh = 0.02 h

        Δh = 0.02 20

        Δh = 0.1 20 cm

        Δh = 0.4 cm = 4 mm

5 0
3 years ago
Aiman is 80 inches tall. How tall is Aiman in centimeters? Show your work.
harkovskaia [24]
203.2 centimeters.
This is because 1 inch = 2.54 centimeters which means multiply 80 inches X 2.54 = 203.2 centimeters.
5 0
2 years ago
In a shear box test on sand a shearing force of 800 psf was applied with normal stress of 1750 psf. Find the major and minor pri
ryzh [129]

Answer:

The major and minor stresses are as 2060.59 psf, -310.59 psf and 1185.59 psf.

Explanation:

The major and minor principal stresses are given as follows:

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}

Here

  • \sigma_x is the normal stress which is 1750 psf
  • \sigma_y is 0
  • \tau_{xy} is the shear stress which is 800 psf

So the formula becomes

\sigma_{max}=\dfrac{\sigma_x+\sigma_y}{2}+\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{max}=\dfrac{1750+0}{2}+\sqrt{\left(\dfrac{1750-0}{2}\right)^2+(800)^2}\\\sigma_{max}=875+\sqrt{\left(875)^2+(800)^2} \\\sigma_{max}=875+\sqrt{765625+640000}\\\sigma_{max}=875+1185.59\\\sigma_{max}=2060.59 \text{psf}

Similarly, the minimum normal stress is given as

\sigma_{min}=\dfrac{\sigma_x+\sigma_y}{2}-\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}\\\sigma_{min}=\dfrac{1700+0}{2}-\sqrt{\left(\dfrac{1700-0}{2}\right)^2+(800)^2}\\\sigma_{min}=875-\sqrt{(875)^2+(800)^2}\\\sigma_{min}=875-\sqrt{765625+640000}\\\sigma_{min}=875-1185.59\\\sigma_{min}=-310.59 \text{ psf}

The maximum shear stress is given as

\tau_{max}=\dfrac{\sigma_{max}-\sigma_{min}}{2}\\\tau_{max}=\dfrac{2060.59-(-310.59)}{2}\\\tau_{max}=\dfrac{2371.18}{2}\\\tau_{max}=1185.59 \text{psf}

5 0
3 years ago
Isormophous phase diagram
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Answer:

Phase diagrams represent the relationship between temperature and the composition of phases present at equilibrium. An isomorphous system is one in which the solid has the same structure for all compositions. The phase diagram shown is the diagram for Cu-Ni, which is an isomorphous alloy system.

Hope it help you friend

6 0
3 years ago
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