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4 years ago

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A cannon ball is fired with an arching trajectory such that at the highest point of the trajectory the cannon ball is traveling

at 98 m/s. If the acceleration of gravity is 9.81 m/s^2, what is the radius of curvature of the cannon balls path at this instant?

1 answer:

ELEN [110]4 years ago

8 0

Answer:

The radius of curvature is 979 meter

Explanation:

We have given velocity of the canon ball v = 98 m/sec

Acceleration due to gravity $g=9.81m/sec^2$

We know that at highest point of trajectory angular acceleration is equal to acceleration due to gravity

Acceleration due to gravity is given by $a_c=\frac{v^2}{r}$ , here v is velocity and r is radius of curvature

So $\frac{98^2}{r}=9.81$

r = 979 meter

So the radius of curvature is 979 meter

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3 years ago

Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent.

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Answer:

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Explanation:

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Mass Balance

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Energy Balance

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$\dot m = \frac{\left(60\,\frac{m}{s} \right)\cdot (0.015\,m^{2})}{0.055665\,\frac{m^{3}}{kg} }$

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b) The exit velocity of steam is:

$\dot m = \frac{v_{out}\cdot A_{out}}{\nu_{out}}$

$v_{out} = \frac{\dot m \cdot \nu_{out}}{A_{out}}$

$v_{out} = \frac{\left(16.168\,\frac{kg}{s} \right)\cdot \left(5.89328\,\frac{m^{3}}{kg} \right)}{0.14\,m^{2}}$

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Answer

given,

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velocity function

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