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vaieri [72.5K]
3 years ago
9

Suppose a star the size of our Sun, but of mass 9.0 times as great, were rotating at a speed of 1.0 revolution every 17 days. If

it were to undergo gravitational collapse to a neutron star of radius 15 km, losing 3/4 of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times
Physics
1 answer:
Tanya [424]3 years ago
4 0
Use the conservation of angular momentum; angular momentum at the beginning = angular momentum at the end 
Conservation of angular momentum: 
I1 w1 = I2 w2 
Where I is the moment of inertia. For a sphere, I=2/5 m R^2. Substituting into the equation above we get 
w2 = I1 w1 / I2 = w1 m1 R1^2 / (m2 R2^2) 
w2 = w1 4 * (R1/R2)^2
= 4*(1)*(7E5/7.5)^2
= 3.48E10 revs/(17days)
= 2.04705882 x 10^9 revs/sec
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Please choose the answer that describes the scientific notation 5098000
bonufazy [111]

So there is a decimal after the last zero and it looks like this 5098000. You have to move the decimal point six back to get in between the five and the zero which looks like this 5.098000 

<span>Scientific notation is the way that scientists easily handle very large numbers or very small numbers. For example, instead of writing 0.0000000056, we write 5.6 x 10^<span>9</span>.</span>

Being that we moved the decimal six places back the answer is 5.098 x 10^6

3 0
3 years ago
A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

4 0
2 years ago
If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

The negative sign implies it's a negative acceleration.

Frictional torque that brought it to rest must be equal to the original torque.

Torqu = I x §

T = 0.33 x 0.33 = 0.11 Nm

5 0
3 years ago
Which properties do metalloids share with metals?
Inga [223]
Metals are not brittle so it can’t be the first one or the third one, both metalloids and metals are shiny so it can’t be the second one. Therefore, it would be the last one because both metalloids and metals are shiny and both are solids at room temperature because it is not a high enough melting point.

ANSWER: Both are shiny and are solid at room temperature.
4 0
3 years ago
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Which of earth sphere have the most mass
DochEvi [55]
The Geosphere which is 99.94% of earths mass
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3 years ago
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