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Elena-2011 [213]
3 years ago
15

the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance

between them doubles?
Physics
1 answer:
antoniya [11.8K]3 years ago
7 0
The equation for gravitational force is:

F = GMm/r^2

Where G is a constant, M is the mass of the larger object, m is the mass of the other object, and r is the distance between the two objects.

So:

1600 = GMm/r^2

Now we can find the force when the distance doubles:

F = GMm/(2r)^2

GMm will remain the same since G is a constant and the masses didn't change, so the only difference between this and the first calculation is the denominator.

By simplifying the denominator, we find:

F = GMm/(4r^2)

Now we can see that the only difference between this and the equation given the original distance is that we are dividing by 4. This means that the force will be one fourth of the original force:

1600/4 = 400

The answer is 400.
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A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu
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Answer:

Part (i) work required to pump the contents to the​ rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the​ cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) =  ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i)  work required to pump the contents to the​ rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the​ cone's rim

Extra work required = 28,914.229 lb ×  5ft = 144571.145 lb.ft

Total work required = (281,913.733 +  144571.145) lb.ft

                                 = 426,484.878 lb.ft

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Which part of the eye is used to see things in high detail?
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Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and t
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Known values; mass per / volume and mass of each atom 

average volume=<span> volume per atom. </span>
<span>density*mass per atom = volume per atom </span>

<span>1cm^3 / 7.87 gram * 9.27 *10^-26 kg /atom* 1000 gram / kg=
</span>=<span>1.178 * 10^-22 cm^3 per atom </span>

<span>Convert that to mass </span>
<span>1 cm^3 * (1 m / 100 cm)^3 </span>
 average volume= <span>1.178 *10^-29 m^3 per  atom </span>

<span>distance: </span>

<span>Volume of a cube is L^3 </span>


<span>so L = (1.178*10^-29)^(1/3) </span>

<span>Half way through the cube is distance 1/2 L. </span>
<span>Second cube is 1/2 L </span>
<span>So distance between the center of 2 cubes = L </span>

=2.28 *10^-10 m 
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3 years ago
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano
Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

\epsilon=\frac{v^2}{2}-\frac{\mu}{r}

Here

  • ε is the specific energy
  • v is the velocity which is given as 2.23 km/s
  • μ is the gravitational constant whose value is 398600
  • r is the distance between earth and the meteorite which is 402,000 km

                         \epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2

Value of specific energy is also given as

\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km

Orbit formula is given as

r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0

Putting values in this equation and solving for e via the quadratic formula gives

ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

r_p=a(e-1)\\

Substituting values gives

r_p=133319 (1.086-1)\\r_p=11465.4 km

Now for estimation of altitude z above earth is given as

z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

r_p=\frac{h^2}{\mu}\frac{1}{1+e}

Rearranging this equation gives

h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s

Now the velocity at perigee is given as

v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

\delta =2 sin^{-1} (\frac{1}{e})

Substituting value in the equation gives

\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08

Aiming radius is given as

\Delta =a \sqrt{e^2-1}

Substituting value in the equation gives

\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0
3 years ago
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