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il63 [147K]
3 years ago
12

1. In the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7.

Physics
1 answer:
timofeeve [1]3 years ago
5 0

Answer: n the laboratory, the life time of a particle moving with speed 2.8 x 10^10 cm\s is found to be 2.5 x 10^-7. Calculate the proper life of the ...

Explanation:

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Explanation:

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8 0
2 years ago
If a sound increases 5 dB, the sound becomes _______ times louder.
viva [34]
Well, I'll try to write the formula in a way that's not confusing,
but I'm afraid it might be slightly confusing anyway.

When you're working with dB, the basic rule is

       A change of 10 dB means either multiplying or dividing by 10 .

     Multiply something by 10  ==>  it increases by 10 dB.
     Divide something by 10   ==>  it decreases by 10 dB.

It turns out that another way to write all of this is . . .

     An increase of 10 dB ===> multiply the original amount by 10¹ 
     An increase of 20 dB ===> multiply the original amount by 10²
     An increase of, say, 7 dB ===> multiply the original amount by 10⁰·⁷

     A decrease of 10 dB ===> multiply the original amount by 10⁻¹
     A decrease of 30 dB ===> multiply the original amount by 10⁻³
     A decrease of, say, 13 dB ===> multiply the original amount by 10⁻¹·³

This question says:  The sound increases by  5 dB .

That means the original 'intensity' or 'power' of the sound
is multiplied by
                           10⁰·⁵  =  √10  =  about 3.162 (rounded) . 

From the choices listed, the closest one is  (c).
3 0
3 years ago
Read 2 more answers
g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7
bogdanovich [222]

Answer:

When they are approaching each other

    f_a = 2228.7 \  Hz

When they are passing  each other

    f_a = 2100Hz

 When they are retreating  from each other

     f_a =  1980.7 Hz

Explanation:

From the question we are told that

     The velocity of car one is  v_1 = 13.0 m/s

      The velocity of car two is  v_2 = 7.22 m/s

     The frequency of sound from car one is  f_e = 2.10 kHz

Generally the speed of sound at normal temperature is  v = 343 m/s

  Now as the cars move relative to each other doppler effect is created and this  can be represented  mathematically  as

              f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]

Where v_s is the velocity of the source of sound

            v_o is the velocity of the observer of the sound

            f_o is the actual frequence

             f_a  is the apparent frequency

Considering the case when they are approaching each other

        f_a = f_o [\frac{v +  v_o}{v -  v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e

Substituting value

            f_a = 2100  [\frac{343 +  7.22}{ 343  -  13} ]

              f_a = 2228.7 \  Hz

Considering the case when they are passing  each other    

At that instant

                  v_o = v_s = 0m/s

                   f_o = f_e

               f_a = f_o [\frac{v }{v } ]

              f_a = f_o

Substituting value

             f_a = 2100Hz

Considering the case when they are retreating  from each other    

                f_a = f_o [\frac{v -  v_o}{v +   v_s} ]

          v_o = v_2  

         v_s = v_1

         f_o = f_e      

Substituting value

         f_a = 2100  [\frac{343 -  7.22}{343 +   13} ]    

          f_a =  1980.7 Hz    

7 0
3 years ago
Joanna claims that a large block of ice will cool a substance more than a small block of ice will at the same temperature. To su
Levart [38]

She puts each block of ice in the same 3000 mL beaker, each with 2000 mL of water at room temperature, and measures the temperature before and after adding ice. Therefore, small blocks of ice will have the same temperature.

Joanna puts two blocks of ice (one larger than the other) into separate cups and fills each with water. She compares the final water temperature of the two cups after each block of ice melts.

Put each block of ice in the same 3000 mL beaker, each at room temperature, put 2000 mL of water in it, and measure the temperature before and after adding ice. This way you keep the water at the same temperature in the beginning, then the temperature changes after you add the ice, giving you a better idea of ​​the final temperature reading.

Learn more about Temperature here brainly.com/question/24746268

#SPJ9

                 

3 0
1 year ago
Which of the following is an action-at-a-distance force? friction tension gravity air resistance
tatuchka [14]

Answer:

Action-at-a-Distance Forces. Frictional Force. Gravitational Force. Tension Force ... The force of gravity on earth is always equal to the weight of the object as ... The friction force is the force exerted by a surface as an object moves across it or ... The force of air resistance is often observed to oppose the motion of an object

Explanation:

8 0
2 years ago
Read 2 more answers
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