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Pani-rosa [81]
3 years ago
15

If the Earth were compressed in such a way that its mass remained the same, but the distance around the equator were just one-ha

lf what it is now, what would be the acceleration due to gravity at the surface of the Earth?
Physics
2 answers:
vovikov84 [41]3 years ago
6 0
If the distance around the equator is reduced by half, then the radius is also reduced by half.

Since the acceleration due to gravity is proportional to 1/(radius²),
the acceleration changes by a factor of 1/(1/2)² = 1/(1/4) = <em>4 </em>.

The acceleration due to gravity ... and also the weight of everything on Earth ...
becomes <em>4 times what it is now</em>.
Neporo4naja [7]3 years ago
4 0
G=GM/r^2
Here are 2 ways to look at this: 
1. Since the r is both inverted and squared on the right while acceleration due to gravity (g) is on the top on the left, any factor applied to r will be inverted and squared as applied to g. In this case, r is multiplied by 1/2, so g will be multiplied by (2/1)^2, which is 4. Assuming your teacher uses 9.8 m/s^2 as the acceleration due to gravity, it would be 39.2 m/s^2. <span>
2. </span>Let r(1) be the radius of the earth, r(2) the smaller radius, g(1) is the acceleration due to gravity on earth and g(2) is the new acceleration due to gravity. In this case, r(2) = (1/2R(1))^2
<span>You can re-arrange the equation for things that will not change on one side and things that are changing on the other. 
This would be gr^2=GM. If GM is not changing, we can say GM = g(1)r(1)^2 = g(2)r(2)^2. Plug in what you know, 9.8*r(1)^2=g(2)*(1/2r(1)^2)
square the 1/2: </span>9.8*r(1)^2=g(2)*1/4r(1)^2
the r(1)^2 cancels out: 9.8=g(2)*1/4
then multiply by 4 on each side: 4*9.8=g(2) = 39.2 m/s^2
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If we know the total energy in a system is 30 J, and we know the PE is 20 J. What is the KE?
svetoff [14.1K]
Can you give more description ??

3 0
3 years ago
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spacesh
Svetradugi [14.3K]

Answer:

The time needed is T  = 16.8 s

Explanation:

From the question we are told that

      The magnitude of the stimulated acceleration due gravity is  a  =  0.5 g

        The diameter of the spaceship is  d =  35m

       

Generally the force acting on the spaceship is  

       F  =  ma

Given that the spaceship is rotating it implies that the force experienced by the occupant is a centripetal force so

      F  = \frac{mv^2}{r}

Thus  

       ma  =  \frac{mv^2}{r}

=>    \frac{v^2}{r}  =  a

      Generally the speed of this spaceship is mathematically represented as

      v =  \frac{2 \pi}{T}

=>    v^2  =   [\frac{2\pi}{T}] ^2

=>     \frac{\frac{4\pi^2 r^2}{T^2} }{r}  = 0.5g

=>       \frac{4 \pi^2 r }{T^2} =  0.5 g

=>         T  = \sqrt{ \frac{4\pi^2 r}{0.5g}}

substituting values

          T  = \sqrt{ \frac{4* (3.142)^2 *(35)}{0.5 * 9.8}}

         T  = 16.8 s

4 0
3 years ago
Read 2 more answers
the length of iron rod at 100 C is 300.36 cm and at 159 C is 300.54 cm.Calculate its length at 0 c and coefficient of linear exp
Ugo [173]

Answer:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

Explanation:

From the question given above, the following data were obtained:

Length (L₁) at 100 °C = 300.36 cm

Temperature 1 (θ₁) = 100 °C

Length (L₂) at 159 °C = 300.54 cm

Temperature 2 (θ₂) = 159 °C

Length (L₀) at 0 °C =?

Coefficient of linear expansion (α) =?

L₁ = L₀ (1 + θ₁α)

300.36 = L₀ (1 + 100α) ....(1)

L₂ = L₀ (1 + θ₂α)

300.54 = L₀ (1 + 159α) ..... (2)

Divide equation (2) by (1)

300.54 / 300.36 = L₀ (1 + 159α) / L₀ (1 + 100α)

1.0006 = (1 + 159α) / (1 + 100α)

Cross multiply

1.0006 (1 + 100α) = (1 + 159α)

1.0006 + 100.06α = 1 + 159α

Collect like terms

1.0006 – 1 = 159α – 100.06α

0.0006 = 58.94α

Divide both side by 58.94

α = 0.0006 / 58.94

α = 1.02×10¯⁵ C¯¹

Substitute the value of α into anything of the equation to obtain L₀. Here we shall use equation (2).

300.54 = L₀ (1 + 159α)

α = 1.02×10¯⁵ C¯¹

300.54 = L₀ (1 + 159 ×1.02×10¯⁵)

300.54 = L₀ (1 + 0.0016218)

300.54 = L₀ (1.0016218)

Divide both side by 1.0016

L₀ = 300.54 / 1.0016

L₀ = 300.05 cm

Summary:

The length at 0 °C is 300.05 cm

Coefficient of linear expansion of iron is 1.02×10¯⁵ C¯¹

6 0
3 years ago
How is energy conserved in a transformation?
irina [24]
As the water plunges, its velocity increases. Its potential energy<span> becomes kinetic</span>energy<span>. The law of conservation of </span>energy<span> states that when one form of </span>energy<span> is</span>transformed<span> to another, no </span>energy<span> is destroyed in the process. ... So the total amount of </span>energy<span> is the same before and after any </span>transformation<span>.

hope it helps

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5 0
3 years ago
4. Provide an example of an isolated energy system and explain how it could be changed to create an open energy system. You may
GarryVolchara [31]
<span>Easy, take the top off your Thermos bottle filled with hot coffee. Assuming perfect insulation, that hot coffee is isolated from the environment; but when the top is opened the heat can now escape to that environment.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
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4 0
3 years ago
Read 2 more answers
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