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natta225 [31]
3 years ago
8

If the mass of a material is 42 grams and the volume of the material is 15 cm^3, what would the density of the material be?

Physics
1 answer:
Fantom [35]3 years ago
3 0

Density = mass / volume

Density = 42g / 15cm^3

Density = 2.8g/cm^3

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An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip
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Answer:

The value is v_2 =  5.53 \  m /s

Explanation:

From the question we are told

  The pipe diameter at location 1 is  d  = 8.8 \  cm =  \frac{8.8 }{10} = 0.88 \ m

   The velocity at location 1 is  v_1 =  2.4 \  m /s

   The diameter at location 2 is  d_2 =  5.80 \  cm  =  0.58 \  m

Generally the area at location 1 is  

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=>     A_1 =  \pi *  \frac{0.88^2}{ 2}

=>     A_1 = 3.142 *  \frac{0.88^2}{ 2}

=>     A_1 = 1.2166 \  m^2

Generally the area at location 1 is  

       A_2 =  \pi *  \frac{d_1^2}{ 2}

=>     A_2 =  \pi *  \frac{0.58^2}{ 2}

=>     A_2 = 0.528  \  m^2

Generally from continuity equation we have that

     A_1 * v_1 =  A_2 * v_2

=>   1.2166 *   2.4   =  0.528   * v_2

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3 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
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