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IgorLugansk [536]
3 years ago
14

State two advantages of thermal expansion​

Physics
1 answer:
Anna71 [15]3 years ago
6 0

Higher Power Efficiency  and Eliminates Risk of Compressor Breakdown

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If object a has more mass than object to be does object contain more matter explain
prohojiy [21]
The object with more mass will contain more matter. Mass is the measurement of the amount of matter an object contains.
7 0
3 years ago
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
NISA [10]

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.

5 0
3 years ago
2. Perform the following:<br>2. 1111+1102<br>​
JulsSmile [24]
2213

just add with a calculator
6 0
3 years ago
Read 2 more answers
A 2kg mass is moving at 3m/s. What is its kinetic energy?
Illusion [34]
<h2><u>KINETIC ENERGY</u></h2>

<h3>Problem:</h3>

» A 2kg mass is moving at 3m/s. What is its kinetic energy?

<h3>Answer:</h3>
  • \color{hotpink} \bold{9 \: J} \\

— — — — — — — — — —

<h3>Formula:</h3>

To calculate the velocity of a kinetic energy, we can use formula

  • \underline{ \boxed{  \tt KE =  \frac{1}{2} m{v}^{2}  } }

where,

  • v is the velocity in m/s
  • KE is the kinetic energy in J (joules)
  • m is the mass in kg

— — —

Based on the problem, the givens are:

  • KE (Kinetic energy) = ? (unknown)
  • m (mass) = 2 kg
  • v (velocity) = 3 m/s

<h3>Solution:</h3>

To get the velocity, substitute the givens in the formula above then solve.

\:   \: \tt KE =  \frac{1}{2} m{v}^{2}   \\ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \: \tt \:   KE =  \frac{1}{2} \times  2 \times {(3)}^{2}  \\ \tt  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  KE =  \frac{1}{2}  \times(2\times 9) \\ \tt  KE =  \underline{ \boxed{ \blue{ \tt9 \: J}}}

Therefore, the kinetic energy is 9 Joules.

3 0
2 years ago
How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
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