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2 years ago

Planet A has a mass of 2.25 x 1020 kg, while planet B has a

Physics

2 answers:

denis23 [38]2 years ago

6 0

The gravitational attraction between two planets is 4905.95 N

<h3>What is gravitational attraction?</h3>

When two objects with masses are placed at a distance, there will an attractive force acting between them.

According to the Newton's law of gravitation, gravitational force is

F = Gm₁m₂ /r²

where r is the distance between the masses m₁ and m₂ and G is the gravitational constant G = 6.67 x 10⁻¹¹ N-m²/kg²

Substitute the values into the expression, we get

F = 6.67 x 10⁻¹¹ x 2.25 x 10²⁰ x 6.20 x 10¹⁸ / (435,500 x 1000)²

F= 4905.95 N

Thus, the gravitational attraction between two planets is 4905.95 N.

Learn more about gravitational attraction.

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attashe74 [19]2 years ago

3 0

Answer:

4905.95 N

Explanation:

F= G m1 m2 / (r^2)

= 6.67 x 10^-11 * 2.25 * 10^20 * 6.2 * 10^10 / ( 435 500 000)^2 =

( be sure to use meters NOT kilometers !)

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What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a

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Answer: d. 5 m/s^2

Explanation:

Acceleration is the change in velocity in a given time.

a = (30-20)/2 = 5

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3 years ago

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Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

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3 years ago

Find the net force produced by a 6 N and an 8 N when acting in same

tankabanditka [31]

Answer:

14 newtons is the answer

Explanation:

because if they act in the same direction they are added together

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2 years ago

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A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00

ira [324]

Answer:

A. 51.42 m.

B. 17.14 s.

Explanation:

Using equations of motion:

vf^2 = vi^2 + 2 * aS

Where,

vf = final velocity

a = acceleration

S = distance to which swan traveled

vi = 0 m/s

6.00^2 = 2 * 0.350S

S = 36/0.7

= 51.42 m.

vf = vi + at

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t = 6/0.35

= 17.14 s.

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3 years ago

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Answer:

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