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VikaD [51]
3 years ago
9

Which of the following will cause an increase in gas pressure in a closed container?

Physics
1 answer:
Shalnov [3]3 years ago
5 0
The answer is A. Or the first option. Pressure is changed by lowering the pressure, not reducing the volume. You would assume its C but its A. 
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Which portion of the electromagnetic spectrum is used in a microscope?
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Answer is C.
 
Optical microscope involves passing visible light transmitted through or reflected from the sample through a single or multiple lenses to allow a magnified view of the sample.

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How does the eye control the amount of light entering it?​
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Answer:

Dilating the pupils.

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A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if
Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long.  The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

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3 years ago
The second law of thermodynamics says that whenever energy is converted from one form to another in a physical or chemical chang
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3 0
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
3 years ago
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