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Elina [12.6K]
3 years ago
5

A horse on the merry-go-round moves according to the equations r = 8 ft, u = (0.6t) rad, and z = (1.5 sin u) ft, where t is in s

econds. Determine the cylindrical components of the velocity and acceleration of the horse when t = 4 s.
Engineering
1 answer:
Luba_88 [7]3 years ago
5 0

Answer:

velocity = 0.664 ft/s, acceleration= 0.365 ft/s²

Explanation:

z= 1.5 sin 0.6t

angular velocity= dz/dt= d(1.5sin0.6t)/dt

                       = 0.9cos0.6t

                    at t=4

                       =0.9cos(0.6×4)

                        = -0.664 ft/s

Acceleration= dv/dt

                      = d(0.9cos0.6t)/dt

                      = -0.54sin0.6t

             at t=4

                   acceleration= -0.54sin(0.6××4)

                                           = 0.365 ft/s²

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Answer:

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Explanation:

Given Data:

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Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.

Thus change in energy is

=(Mass * change in P)/density\\= \frac{M*P}{p}\\\\

As we know that Mass = Volume x density

substituting the value

Energy = Volume * density x ΔP / density

Change in energy = Volumetric flow x ΔP

Change in energy = 0.226 x 8.274 = 1.869 KW

Now mechanical efficiency = change in energy / work done by shaft

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2 years ago
Q9. A cylindrical specimen of a metal alloy 54.8 mm long and 10.8 mm in diameter is stressed in tension. A true stress of 365 MP
wolverine [178]

Answer:

σ = 391.2 MPa

Explanation:

The relation between true stress and true strain is given as:

σ = k εⁿ

where,

σ = true stress = 365 MPa

k = constant

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ε = (61.8 - 54.8)/54.8 = 0.128

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Therefore,

365 MPa = K (0.128)^0.2

K = 365 MPa/(0.128)^0.2

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Now, we have the following data:

σ = true stress = ?

k = constant = 550.62 MPa

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ε = (64.7 - 54.8)/54.8 = 0.181

n = strain hardening exponent = 0.2

Therefore,

σ = (550.62 MPa)(0.181)^0.2

<u>σ = 391.2 MPa</u>

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