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adell [148]
3 years ago
14

In a classroom demonstration, the pressure inside a soft drink can is suddenly reduced to essentially zero. You may want to revi

ew (Pages 508 - 512) .Assuming the can to be a cylinder with a height of 11 cm and a diameter of 6.8 cm , find the net upward force exerted on the vertical sides of the can due to atmospheric pressure.?

Physics
1 answer:
KengaRu [80]3 years ago
4 0

Answer:

Net upward force = 2373.72N

Explanation:

The concept of Pressure = Force/Area is applied as the steps are shown in the attachment

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Which is the best description of a high pressure system?
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Answer:

The Answer is gonna be B.

Air warms and rises.

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3 years ago
During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel?
SIZIF [17.4K]

-70560, -9.8/2 = -4.9

120^2 = 14400.

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4 years ago
The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the
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The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.

The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>

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3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
ruslelena [56]

Answer:

a

The orbital speed is v= 2.6*10^{3} m/s

b

The escape velocity of the rocket is  v_e= 3.72 *10^3 m/s

Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
4 years ago
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