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Bess [88]
3 years ago
9

When you are standing on Earth, orbiting the Sun, and looking at a broken cell phone on the ground, there are gravitational pull

s on the cell phone from you, the Earth, and the Sun. Rank the gravitational forces on the phone from largest to smallest. Assume the Sun is roughly 109 times further away from the phone than you are, and 1028 times more massive than you. Rank the following choices in order from largest gravitational pull on the phone to smallest. To rank items as equivalent, overlap them.
a. Pull phone from you
b. Pull on phone from earth
c. Pull on phone from sun
Physics
1 answer:
Mandarinka [93]3 years ago
5 0

Answer:

The answer is "Option b, c, and a".

Explanation:

Here that the earth pulls on the phone, as it will accelerate towards Earth when we drop it.

We now understand the effects of gravity:

F \propto  M\\\\F\propto  \frac{1}{r^2}\\\\or\\\\F \propto  \frac{M}{r^2}\\\\Sun (\frac{M}{r^2}) = \frac{10^{28}}{(10^9)^2} = 10^{10}

The force of the sun is, therefore, 10^{10} times greater and the proper sequence, therefore, option steps are:

b. Pull-on phone from earth

c. Pull-on phone from sun

a. Pull phone from you

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if a company mistakenly counts more items during a physical inventory than actually exist, how will the error affect their botto
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Answer:

The cost of the merchandise was not included in ending inventory.

Net income for the current year will be understated.

Explanation:

4 0
3 years ago
A parallel circuit has two 8.0-ohm resistors and a power source of 9.0 volts. If a 12.5-ohm resistor is added to the circuit in
Airida [17]
-- The two 8-ohm resistors in parallel are equivalent to
a single 4-ohm resistor.

-- The current in the circuit is 

                 (voltage) / (resistance) = (9v) / (4ohms) = 2.25 Amperes

-- If another resistor is added in parallel, then no matter how large
or small it is, the current in the circuit increases. 
(When another lane is added to a road, the traffic capacity of the
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The circuit is now equivalent to a 12.5-ohm resistor in parallel
with 4-ohms.  That's ...

       (12.5 x 4) / (12.5 + 4)  =  50/16.5  =  3.03 ohms .

-- The new current is  (9v) / (3.03 ohms)  =  2.97 Amperes
            
7 0
3 years ago
A hot water stream at 80 oC enters a mixing chamber with mass flow rate of 3.6 kg/s and mixed with cold water at 20 oC. If the m
poizon [28]

Explanation:

The mixing chamber will be well insulated when steady operating conditions exist such that there will be negligible heat loss to the surroundings. Therefore, changes in the kinetic and potential energies of the fluid streams will be negligible and there are constant fluid properties with no work interactions.

   T < T_{sat} at 250 kPa = 127.41^{o}C

   h_{1} approx equal to h_{f} at 80^{o}C

              = 335.02 kJ/kg

    h_{2} ≈ h_{f} at 20^{o}C

                      = 83.915 kJ/kg

and,    h_{3} ≈ h_{f} at 42^{o}C = 175.90 kJ/kg

Therefore, mass balance will be calculated as follows.

   m^{o}_{in} - m^{o}_{out} = \Delta m^{o}_{system} \rightarrow m^{o}_{1} + m^{o}_{2} = m^{o}_{3}

And, energy balance will be given as follows.

      E^{o}_{in} - E^{o}_{out} = \Delta E^{o}_{system}

As we are stating steady conditions,

     \Delta m^{o}_{system} and \Delta E^{o}_{system} cancel out to zero.

So,    E^{o}_{in} = E^{o}_{out}

     m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = m^{o}_{3}(h_{3})

On combining the relations, we solve for m^{o}_{2} as follows.

   m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = (m^{o}_{1} + m^{o}_{2})(h_{3})

   m^{o}_{2} = (\frac{(h_{1} - h_{3})}{(h_{3} - h_{2})}) \times m^{o}_{1}

              = \frac{(335.02 - 175.90)}{(175.90 - 83.915)} \times 0.5  

       m^{o}_{2} = 0.865 kg/s

                       = 51.9 kg/min      (as 1 min = 60 sec)

Thus, we can conclude that the mass flow rate of cold stream is 51.9 kg/min.

6 0
3 years ago
Pllllllllllllllllllssssssssss help me will give brainiest if correct
Gelneren [198K]

Answer:

I believe it is the same as before-Electrical energy is transformed into heat and light energy.

Explanation:

Hope this helps'-'

8 0
3 years ago
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