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irinina [24]
3 years ago
6

Look at the densities of the jovian planets given in figure 1. which of the following statements best describes the pattern of j

ovian planet densities?
Physics
2 answers:
barxatty [35]3 years ago
4 0

more than 5.5 times the density of water

is the answer.

avanturin [10]3 years ago
3 0
Given the densities of the jovian planets, the densities is just like random numbers, so the densities of the jovian planets have no concrete pattern due to its randomness. pattern can be either linear or geometric, where linear you see a common difference among data, while geometric you observed a common ratio
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A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

\rho is the density of the flow

V = Velocity

C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

m= 1600lbf = 725.747Kg

\rho = 1.21 kg/m^3

C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

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Explanation:

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umka2103 [35]

Answer:

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Explanation:

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I actually know the answer to this one, you use pennies to find the atomic weight of a penny, it really doesn't have a weight. LOL

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3 years ago
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(4.56 x 10^-13)-(1.17 x 10^-13)
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3.39 x 10^-13

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