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azamat
2 years ago
11

a tank initially holds 100 gallons of a brine solution containing 1lb of salt. at t=0 another brine solution containing 1 lb of

salt per gal is poured into the tank at the rate of 3 gal/min, while the well-stirred mixture leaves the tank at the rate. find the amount of salt in the tank at any time t.
SAT
1 answer:
vaieri [72.5K]2 years ago
8 0

Answer:

dodo

Explanation:

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By considering the conservation of energy, the speed V of the spacecraft when it eventually crash into the earth and  its distance from the center of the earth is r are 11172.6 m/s and 3294.6 m/s respectively.

Given that, r is at infinity when a spacecraft has run out of fuel and its kinetic energy is zero. if only the gravitational force of the earth were to act on the spacecraft would eventually crash into the earth.

Where

Mass of the earth  m_{e} = 5.97 × 10^{24} kg

Radius of the earth R_{e} = 6.38 × 10^{6} m

G = 6.67 x 10^{-11}Nm^{2}kg^{-2}

a.)  Since the spacecraft is primarily moving through the near vacuum of space, the speed V of the spacecraft when it eventually crash into the earth can be calculated by considering the conservation of energy.

The total initial energy  =  total final energy.

E_{i}  =  E_{f}

K.E_{i} + U_{i} = K.E_{f} + U_{f}

where K.E_{i} = 0

-Gm_{e}m/(r + R_{e}) = 1/2mV^{2} + ( -Gm_{e}m)/R_{e}

mass of the spacecraft will cancel out

Gm_{e}/(r + R_{e}) = 1/2V^{2} + ( -Gm_{e})/R_{e}

Gm_{e}/(r + R_{e}) = 0 since r is at infinity.

0 =  1/2V^{2} + ( -Gm_{e})/R_{e}

Substitute all the parameters into the equation above.

0 = 0.5V^{2} + ( -6.67 x 10^{-11} x 5.97 x 10^{24} )/6.38 x 10^{6}

0.5V^{2} = 62,413,636.36

V^{2} =  62,413,636.36 / 0.5

V^{2} = 124,827,272.7

V = \sqrt{124,827,272.7}

V = 11172.6 m/s

b.) Given that the distance from the earth r = x(R_{e})

That is,

r = 11.5 x 6.38 x 10^{6}

r = 73,370,000 m

To find the spacecraft's speed when its distance from the center of the earth is r, we will use the same formula. That is,

K.E_{i} + U_{i} = K.E_{f} + U_{f}

0 =  1/2V^{2} + ( -Gm_{e})/r

Substitute all the parameters into the formula

0 = 0.5V^{2} + ( -6.67 x 10^{-11} x 5.97 x 10^{24} )/ 73,370,000

0.5V^{2} = 5427272.727

V^{2} = 10854545.45

V = \sqrt{10854545.45}

V = 3294.6 m/s

Therefore, the speed V of the spacecraft when it eventually crash into the earth and  its distance from the center of the earth is r are 11172.6 m/s and

3294.6 m/s respectively.

Learn more about gravitation potential here: brainly.com/question/940770

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