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34kurt
2 years ago
9

Compute the acceleration of gravity for a given distance from the earth's center, distCenter, assigning the result to accelGravi

ty. The expression for the acceleration of gravity is: (G * M) / (d2), where G is the gravitational constant 6.673 x 10-11, M is the mass of the earth 5.98 x 1024 (in kg) and d is the distance in meters from the earth's center (stored in variable distCenter).Sample program:#include int main(void) { const double G = 6.673e-11; const double M = 5.98e24; double accelGravity = 0.0; double distCenter = 0.0; distCenter = 6.38e6; printf("accelGravity: %lf\n", accelGravity); return 0;
Engineering
2 answers:
Elodia [21]2 years ago
7 0

Answer:

See Explaination

Explanation:

#include <stdio.h>

int main(void)

{

const double G = 6.673e-11;

const double M = 5.98e24;

double accelGravity = 0.0;

double distCenter = 0.0;

distCenter = 6.38e6;

//<StudentCode>

accelGravity = (G * M) / (distCenter * distCenter);

printf("accelGravity: %lf\n", accelGravity);

return 0;

}

saveliy_v [14]2 years ago
5 0

Answer:

accel_gravity = (G * M) / dist_center **2

Explanation:

They give it to you in code so you can figure it out, but its right in front of you.

They give you, The expression for the acceleration of gravity is: (G * M) / (d 2).

All you have to do is code it.

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500 =  (T - 20)\frac{4\pi \times0.52\times 1\times 10}{10 - 1}

T = 68.86 +20 = 88.865^{\circ}C  

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2 years ago
A shaft consisting of a steel tube of 50-mm outer diameter is to transmit 100 kW of power while rotating at a frequency of 34 Hz
nikitadnepr [17]

Answer:

25 - \sqrt[4]{26.66*10^{-8} }  mm

Explanation:

Given data

steel tube : outer diameter = 50-mm

power transmitted = 100 KW

frequency(f) = 34 Hz

shearing stress ≤ 60 MPa

Determine tube thickness

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power = T(torque) * w (angular velocity)

angular velocity ( w ) = 2\pif = 2 * \pi * 34 = 213.71

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next we calculate the inner diameter  using the relation

  \frac{J}{c_{2}  } = \frac{T}{t_{max} }  = 467.92 / (60 * 10^6) =  7.8 * 10^-6 m^3

also

c2 = (50/2) = 25 mm

\frac{J}{c_{2} } = \frac{\pi }{2c_{2} } ( c^{4} _{2} - c^{4} _{1} ) =  \frac{\pi }{0.050} [ ( 0.025^{4} - c^{4} _{1}  ) ]

therefore; 0.025^4 - c^{4} _{1} = 0.050 / \pi (7.8 *10^-6)

c^{4} _{1} = 39.06 * 10 ^-8 - ( 1.59*10^-2 * 7.8*10^-6)

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c_{1} = \sqrt[4]{26.66 * 10^{-8} }  =

THE TUBE THICKNESS

c_{2} - c_{1} = 25 - \sqrt[4]{26.66*10^{-8} }  mm

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Answer:

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