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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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Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
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\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
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3 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

4 0
3 years ago
Describe the energy change in the particles of a substance during melting. A) The kinetic energy of the particles remains unchan
Rina8888 [55]

Answer:

As ice melts into water, kinetic energy is being added to the particles. This causes them to be 'excited' and they break the bonds that hold them together as a solid, resulting in a change of state: solid -> liquid.

Explanation:

As we may know, the change in state of an object is due to the change in the average kinetic energy of the particles.

This average kinetic energy is proportional to the temperature of the particles.

This is because heat is a form of energy; by adding energy to ice - heat, you "excite" the water molecules, breaking the interactions in the lattice structure and forming weaker, looser hydrogen-bonding interactions.

This causes the ice to melt. This is demonstrated in the image below.

More generally, when you remove energy - the object cools down, the particles move a lot slower. So slow, that they individually attract other molecules more than before, and this results in a physical change that also changes the state.

3 0
3 years ago
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Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk.
stellarik [79]

Answer: The correct option is Option b.

Explanation:

Power is defined as the rate of work done by an object.

Mathematically,

P=\frac{W}{t}    .....(1)

And work done is the product of force exerted on the object times the displacement covered by that object.

Mathematically,

W=F.s

Putting this value in above equation, we get:

P=\frac{F.s}{t}

where,

P = power = ?W

F = Force exerted = 10N

s = Displacement = 400cm = 4m   (Conversion factor: 1m = 100 cm)

t = Time taken = 8s

Putting values in above equation, we get

P=\frac{10\times 4}{8}\\\\P=5W

Hence, the correct option is Option b.

7 0
3 years ago
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What is buoyant force?
Nostrana [21]

This is the upthrust on an object which is placed inside a fluid

This force act upwards and always push upwards

so the correct answer is given as

D. A force within a fluid that pushes upward

this force is always due to pressure difference at two levels of

at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy

3 0
3 years ago
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