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Gennadij [26K]
3 years ago
10

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

in the second trial 4H. Compare the FINAL VELOCITY'S for the packages right as they hit the
ground (or right before)?
Physics
1 answer:
enyata [817]3 years ago
5 0

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

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The pitch of a sound wave refers to the loudness of a wave. True False
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B. How can you tell where sugar enters the blood?
Korolek [52]

Answer:

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Explanation:

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I hope I helped

8 0
3 years ago
A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res
Natalija [7]

Answer:

The time constant is \tau = 17.5 \ s    

Explanation:

From the question we are told that

   The spring constant is  k = 11.5 \  N/m

   The mass  of the ball is  m_b  = 490 \ g  = 0.49 \ kg

   The amplitude of the  oscillation t the beginning is x =  6.70 cm = 0.067 \  m

    The amplitude after time t is  x_t = 2.20 cm = 0.022 \  m

    The number of oscillation is N  = 30

Generally the time taken to attain the second amplitude is mathematically represented as

       t  = N  *  T                                            Here  T is the period of oscillation

         t = N * 2\pi \sqrt{\frac{m}{k} }

=>     t = 30 * 2 * 3.142 *  \sqrt{\frac{ 0.490}{11.5} }

=>     t = 38.88 \  s

Generally the amplitude at time t is mathematically represented as

         x(t) = x e^{-\frac{at}{2m} }

Here a is the damping  constant so

 at  t = 38.88 \  s ,  x_t = 2.20 cm = 0.022 \  m

So  

     0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }

=>  0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }

taking natural log of both sides

=>  ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }    

=>   a = 0.028

Generally the time constant is mathematically represented as

    \tau = \frac{m}{a}      

=> \tau = \frac{0.490}{  0.028}    

=> \tau = 17.5 \ s    

4 0
3 years ago
Imagine that two balls, a basketball and a much larger exercise ball, are dropped from a parking garage. If both the mass and ra
pashok25 [27]

Here if we assume that there is no air friction on both balls then we can say

F = mg

now the acceleration is given as

F = ma = mg

a = g

so here both the balls will have same acceleration irrespective of size and mass

so we can say that to find out the time of fall of ball we can use

y = \frac{1}{2}gt^2

t = \sqrt{\frac{2y}{g}}

now from above equation we can say that time taken to hit the ground will be same for both balls and it is irrespective of its mass and size

3 0
3 years ago
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