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garri49 [273]
3 years ago
12

Mars has a mass of about 6.58 × 1023 kg, and its moon Phobos has a mass of about 9.3 × 1015 kg. If the magnitude of the gravitat

ional force between the two bodies is 4.18 × 1015 N, how far apart are Mars and Phobos? The value of the universal gravitational constant is 6.673 × 10−11 N · m2 /kg2 . Answer in units of m.
Physics
1 answer:
NISA [10]3 years ago
5 0

This looks complicated, but it's actually not too tough.

The formula for the gravitational force between two objects is

              Force = G  (one mass) (other mass) / (distance²) .

The question GAVE us all of those numbers except the distance.
All we have to do is pluggum in, massage it around, and find
the distance. 

Force  =         4.18 x 10¹⁵     N
  G  =             6.673 x 10⁻¹¹  N·m²/kg²
One mass =   6.58 x 10²³     kg
Other mass = 9.3 x 10¹⁵       kg   .

The only tricky thing about this is gonna be the arithmetic ...
keeping all the exponents straight.

Take the formula for the gravitational force and plug in
everything we know:

Force = (G) · (one mass) · (other mass) / (distance²) 

4.18x10¹⁵N = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (distance²).

Multiply each side by  (distance²):

(distance²)·(4.18x10¹⁵N) = (6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) 

Divide each side by  (4.18 x 10¹⁵ N) :

(distance²)=(6.673x10⁻¹¹N-m²/kg²)·(6.58x10²³kg)·(9.3x10¹⁵kg) / (4.18x10¹⁵N)

That's the end of the Physics and Algebra.  The only thing left is Arithmetic.
We have to simplify that whole ugly thing on the right side of the equation,
and then take the square root of each side.

When I crunch down the right side of that equation, I get

           (distance²)  =  9.769 x 10¹³  m²

and when I take the square root of each side, I get

             distance  =  9.884 x 10⁶ meters .   **

You should check my Arithmetic.   **
(Pause occasionally to let your calculator cool off.)


BY THE WAY ... 
That "distance" in the equation for gravitational force is the distance
between the CENTERS of the two objects. 
This doesn't make much difference for Phobos, because Phobos isn't
much bigger than a big sweet potato.  But it does make a difference for
Mars. 
The 'distance' we find with all of this nonsense is NOT the distance
between Phobos and the surface of Mars.  It's the distance between 
Phobos and the CENTER of Mars, so it includes the planet's radius.   


** Consulting online resources between Floogle and Flickerpedia,
I found that the orbital distance of Phobos from Mars varies between
9,234 km and 9,517 km.  Add the planet's radius to these, and I'm
beginning to feel confidence in the results of my back-of-the-napkin
calculation.  But you should still check my Arithmetic.

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1. Two vectors have magnitudes 6 & 8 units, respectively. Find the maximum &
defon

Answer:

<em>The maximum magnitude of their resultant is 14 units, and the minimum is 2 units</em>

Explanation:

<u>Resultant Vectors</u>

When two or more vectors are added or subtracted, the resultant vector can be found by considering their magnitudes and directions.

Two vectors applied to the same point can produce a result that can vary from being completely collaborative or completely opposite.

If two vectors act in a collaborative form, their magnitudes are added and the result has the maximum possible magnitude. If they act in opposite directions, the result has the minimum possible magnitude.

Thus, being 6 and 8 units the magnitudes of the vectors, the maximum magnitude of their resultant is 14 units, and the minimum is 2 units.

5 0
2 years ago
Sketch the electric field around these two objects if they have the same sign of charge. Make a separate drawing showing equipot
diamong [38]

Answer:

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

Explanation:

The electric field around two objects with charge of the same sign, for simplicity suppose that the objects have positive point spherical charges,

          E = k q / r2

bold letters indicate vectors, therefore the total electric field is

           E_total = E1 + E2

the module of this field is

           E_total = E1- E2

therefore we can outline this field

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

An outline of these shows in Attachment A

The equipotential surfaces are defined as being perpendicular to the electric field lines since the electric field and the power difference are related

              E = \frac{dV}{dx} i^ + \frac{dV}{dy} j^ + \frac{dV}{dz} k^ = \Delta V

We can schematize some characteristics of these surfaces

* very close to each load are spherical surfaces

* very far from the load is an elliptical surface, which envelops the loads

* between them there is a point of zero potential point C

See attached part B

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3 years ago
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Ira Lisetskai [31]

I believe the term Frequency is what you are looking for.  

8 0
3 years ago
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8. When a 100 N bag of nails hangs motionless from a single vertical strand of rope, how many newtons of tension are exerted in
Svetllana [295]

If the bag is motionless, then it's not accelerating up or down.
That fact right there tells you that the net vertical force on it
is zero.  So the sum of any upward forces on it is exactly equal
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If the bag is suspended from a single rope, then the tension
in the rope must be equal to the 100-N weight of the bag.

And if there are four ropes holding it up, then the sum of
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8 0
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