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son4ous [18]
1 year ago
10

a ball is projected horizontally from the top the building.One second later another ball is projected horizontally from the same

point with the same velocity.At what point In the motion will the balls be closest to each other?Will the first ball always be travelling faster than the second ball?What will be the time interval between them when the ball hits the ground?Can the horizontal projection velocity of the second ball be changed so that the balls arrive at the ground at the same time?
Physics
1 answer:
stellarik [79]1 year ago
8 0

The point In the motion that the balls will be closest to each other is as at the time as the second ball is thrown or lunched.

<h3>Will the first ball always be travelling faster than the second ball?</h3>

No, The two motions are said to be parallel to one another, so they that they both the same timing. Hence, the time it takes for both balls to be able to fall to the ground is said to be the same.

Yes, the both balls have the same speed because of  two balls are said to hit the ground, they are said to be done at the same height as well as the same energy and so they have the same speed.

Note that balls will take about 2 seconds to be able to hit the ground.

The horizontal projection velocity of the second ball can be changed so that the balls arrive at the ground at the same time and so the balls will be closest to one another if the second ball is thrown.

Learn more about projection velocity  from

brainly.com/question/4582453

#SPJ1

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Answer:

The frequency shift detected is \Delta  f  =1010.3 Hz

Explanation:

From the question we are told that

    The normal frequency of the state highway patrol radar guns f = 8.66 GHz =   8.66 *10^{9} \ Hz

     The speed of approach is v  =  35 .0 \ m/s

       

Now the frequency measure by the the state highway patrol radar guns as your car approaches is mathematically represented as

        f_n  =  f (1 + \frac{v}{c} )

Where c is the speed of light which a has constant value of

     c =  3.0 *10^{8 } \ m/s

  Now

       f_n - f  =  \frac{fv}{c} )

=>    \Delta  f  =   \frac{fv}{c}

substituting values

        \Delta  f  = \frac{ 8.66 *10^{9} *  35}{3.0 *10^8 }

        \Delta  f  =1010.3 Hz

 

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3 years ago
The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan
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To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

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The area of a sphere is given by

A = 4\pi r^2

So replacing we have to

I = \frac{P}{4\pi r^2}

Since the question tells us to find the proportion when

r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}

So considering the two intensities we have to

I_1 = \frac{P_1}{4\pi r_1^2}

I_2 = \frac{P_2}{4\pi r_2^2}

The ratio between the two intensities would be

\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}

The power does not change therefore it remains constant, which allows summarizing the expression to

\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2

Re-arrange to find I_2

I_2 = I_1 (\frac{r_1}{r_2})^2

I_2 = 100*(\frac{1}{5})^2

I_2 = 4W/m^2

Therefore the intensity at five times this distance from the source is 4W/m^2

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Explanation:

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