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3 years ago

PLZ HELP ASAP Saturn and Jupiter both complete one rotation in about____hours.

Physics

2 answers:

Debora [2.8K]3 years ago

8 0

Answer:

Explanation: Earth takes 24 hours to complete one spin, and Mars takes 25 hours. The gas giants rotate really fast. Jupiter takes just 10 hours to complete one rotation. Saturn takes 11 hours, Uranus takes 17 hours, and Neptune takes 16 hours.

grigory [225]3 years ago

6 0

Answer:

10 hours

Explanation:

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The mass of a copper atom is 1.06 10-25 kg, and the density of copper is 8 920 kg/m3 . (a) determine the number of atoms in 1 cm

Andreyy89

Mass of copper atom is $1.06\times 10^{-25}kg$ and density of copper atom is $8920 kg/m^{3}$ .

First calculate the volume from density and mass as follows:

$V=\frac{m}{d}=\frac{1.06\times 10^{-25} kg}{8920 kg/m^{3}}=1.188\times 10^{-29}m^{3}$

Now, convert $m^{3}$ to $cm^{3}$ ,

$1 m^{3} = 10^{6} cm^{3}$

Thus,

$1.188\times 10^{-29}m^{3}=1.188\times 10^{-23}cm^{3}$

Now,

$1.188\times 10^{-23}cm^{3}\rightarrow 1.06\times 10^{-25}kg$

Thus,

$1cm^{3}\rightarrow \frac{1.06\times 10^{-25}kg}{1.188\times 10^{-23}kg}=0.00892 kg$

or,

m=8.92 g

The molar mass of copper is 63.54 g/mol thus, number of moles can be calculated as:

$n=\frac{m}{M}=\frac{8.92 g}{63.54 g/mol}=0.1403 mol$

Now, in 1 mol of Cu there are $6.023\times 10^{23}$ atoms, number of atoms in 0.1403 mol will be:

$N=0.1403\times 6.023\times 10^{23}=8.45\times 10^{22} atoms$

8 0

3 years ago

Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is