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lisabon 2012 [21]
3 years ago
11

Weather balloons appear only partially inflated when launched into the air. Why would scientists do this?

Physics
1 answer:
zvonat [6]3 years ago
7 0

The reason weather balloons are only partially inflated is to allow room for the helium in the balloon to expand.

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A dry cell does 7.5 j of work through chemical energy transfer 5.00C between terminals of the cell . What is the electric potent
lisov135 [29]
Here's a useful factoid that you don't hear about very often:

1 volt means 1 Joule per Coulomb.

When 1 coulomb of charge falls or gets lifted through 1 volt potential difference, it gains or loses 1 Joule of energy.

If you want to lift 5 coulombs to a height of 1 volt, you have to give it 5 joules.

If you actually give those 5 coulombs 7.5 joules instead, they'll rise up to 1.5 volts above the potential where they started. The flowed through a potential DIFFERENCE of 1.5 volts.

(If they started at a point that's connected to the Earth, like a water pipe or a metal flagpole, then their new potential is 1.5 volts, because we define zero as the potential of the ground.)
7 0
3 years ago
To drive a car at a constant velocity, you
kipiarov [429]

Answer:

the answer is C

Explanation:

The car, first is at rest and if you don't accelerate it won't move. When to hit the gas it will accelerate from rest

8 0
2 years ago
A
dangina [55]

Answer:

C 2000v its obviously ans because if o is 1000 2 vo is 2000v

7 0
3 years ago
the average american receives 2.28 mSv dose equivalent from radon each year. Assuming you receive this dose, and it all comes fr
Dmitry [639]

Answer:

The approximate number of decays  this represent  is  N= 23*10^{10}  

Explanation:

 From the question we are told that

    The amount of Radiation received by an average american is I_a = 2.28 \ mSv

     The source of the radiation is S = 5.49 MeV \ alpha \ particle

 Generally

            1 \  J/kg = 1000 mSv

   Therefore  2.28 \ mSv = \frac{2.28}{1000} = 2.28 *10^{-3} J/kg

Also  1eV = 1.602 *10^{-19}J

  Therefore  2.28*10^{-3} \frac{J}{kg} = 2.28*10^{-3} \frac{J}{kg}  * \frac{1ev}{1.602*10^{-19} J} = 1.43*10^{16} ev/kg

           An Average american weighs 88.7 kg

      The total energy received is mathematically evaluated as

        1 kg ------> 1.423*10^{16}ev \\88.7kg  --------> x

Cross-multiplying and making x the subject

           x = 88.7 * 1.423*10^{16} eV

              x = 126.2*10^{16}eV

Therefore the total  energy  deposited is x = 126.2*10^{16}eV

The approximate number of decays  this represent  is mathematically evaluated as

            N = \frac{x}{S}

Where n is the approximate number of decay

   Substituting values

                             N = \frac{126 .2*10^{16}}{5.49*10^6}  

                                  N= 23*10^{10}  

                     

             

7 0
4 years ago
Help me with this review question please.
QveST [7]

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

6 0
3 years ago
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