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3 years ago

15

A car moves along a circular track of radius 250 ft such that its seed for a short period of time 0<_ t <_ 4s, is v = 3(t

- t^2) ft/s, where t is in seconds.
a. Determine the magnitude of its acceleration when t = 3s.

b. How far has it traveled in t = 3s?

1 answer:

beks73 [17]3 years ago

7 0

Answer: a) -5 ft/s²

B) 4.5 ft

Explanation: Radius= 250ft

Velocity V = 3(t-t²)ft/s

A). When t= 3 s, the acceleration is

dv/dt = 1-2t

dv/dt = 1- 2(3)

= 1-6

= -5 ft/s²

B. How far it traveled in 3 sec

Distance= 3t²/2 -t³/3 ft

Substituting 3

Distance = 27/2 - 9

= 13.5 - 9

= 4.5 ft

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For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta

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The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

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Where R is the gas's particular constant defined in terms of Cp and Cv:

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The gamma undimensional constant is also expressed as a function of Cv and Cp:

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$\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}$

$\gamma=1.29$

And the variable T is the temperature in Kelvin. Thus for the known temperature:

$c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)$

$c=\sqrt(91867.73 \frac{J}{Kg})$

The Jules unit can expressing by:

$J=N.m=\frac{Kg.m}{s^2}* m$

$J=\frac{Kg.m^2}{s^2}$

Replacing the new units for the speed of the sound:

$c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})$

$c=\sqrt(91867.73 \frac{m^2}{s^2})$

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For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration will raise the carbon concentratio

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This question is incomplete, the complete question is;

For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.

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