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Elza [17]
4 years ago
7

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

had a mass of 110 kg, how much would you weigh when standing on the surface of this asteroid? Express your answer to two significant figures and include appropriate units. 2. If you could walk on the surface of this asteroid, what minimum speed would you need to launch yourself into an orbit just above the surface of the asteroid? Express your answer to two significant figures and include appropriate units.
Physics
1 answer:
Law Incorporation [45]4 years ago
3 0

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

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Speed= (frequency) x (wavelength)

= (2 per second) x (5 cm) = <em><u>10 cm/sec .</u></em>
6 0
3 years ago
If you were in charge of designing a wire to carry electricity across your city, state or province, which of
Anon25 [30]

Answer:

Thin, aluminium and buried underground.

Explanation:

When it comes to electrification of a state or province, some characteristics of the wire to use must be considered. This would help to minimize and avoid power loss and wire burns.

i. The wire to use should be thin, and a quite number can be twisted one against the other so as to increase the surface area for heat dissipation.

ii. Aluminium wire is more preferable for this project. It has a high melting point, and reduces energy loss.

iii. Burying the wire underground through an insulator is the best choice, though expensive but would preserve the wire from external influence.

7 0
3 years ago
The path of a projectile fired at a 30° angle to the horizontal best described as ?
STALIN [3.7K]
Linear.

Since it is fired at a 30-degree angle, it is assumed to follow a straight line, which makes it linear.
6 0
4 years ago
A number line goes from negative 5 to positive 5. Point D is at negative 4 and point E is at positive 5. A line is drawn from po
saul85 [17]

The location of the point F that partitions a line segment from D to E (\overline{DE}), that goes from <u>negative 4</u> to <u>positive 5,</u> into a 5:6 ratio is fifteen halves (option 4).  

We need to calculate the segment of the line DE to find the location of point F.

Since point D is located at <u>negative -4</u> and point E is at <u>positive 5</u>, we have:

\overline{DE} = E - D = 5 - (-4) = 9

Hence, the segment of the line DE (\overline{DE}) is 9.

Knowing that point F partitions the line segment from D to E (\overline{DE}) into a <u>5:6 ratio</u>, its location would be:

F = \frac{5}{6}\overline{DE} = \frac{5}{6}9 = 5*\frac{3}{2} = \frac{15}{2}  

Therefore, the location of point F is fifteen halves (option 4).

Learn more about segments here:

  • brainly.com/question/24472171?referrer=searchResults
  • brainly.com/question/13270900?referrer=searchResults

       

I hope it helps you!

5 0
3 years ago
Read 2 more answers
A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. When the temperature of th
Nesterboy [21]

Answer:

Option C is the correct answer.

Explanation:

By Charles's law we have

        V ∝ T

That is

       \frac{V_1}{T_1}=\frac{V_2}{T_2}

Here given that

      V₁ = 0.20 cubic meter

      T₁ = 333 K

      T₂ = 533 K

Substituting

      \frac{V_1}{T_1}=\frac{V_2}{T_2}\\\\\frac{0.20}{333}=\frac{V_2}{533}\\\\V_2=\frac{0.20}{333}\times 533=0.3198m^3

New volume of the gas  = 0.3198 m³

Option C is the correct answer.

7 0
4 years ago
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