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netineya [11]
3 years ago
7

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the gr

ound. When the balls collide, they are moving n opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur? (Use any variable or symbol stated above as necessary.) Both balls experience constant acceleration once they are �n flight. Choose a coordinate system with the origin at the ground and the upward direction positive. When the balls collide they are at the same height above the ground.
Physics
1 answer:
kramer3 years ago
3 0

Answer:

2h / 3

Explanation:

Let after time t , collision occurs

distance covered by A

= 1/2gt²

Let u be the initial velocity with which B was thrown upwards

distance by B

= ut - 1/2 gt²

total distance by A and B

h = ut - 1/2 gt² +  1/2gt²

h = ut

t = h / u

velocity of A after t

= gt

= g x h/u

velocity of B after t

= u - gt

= u - g x h/u

given

g x h/u = 2 (u - g x h/u)

2u² - 2gh = gh

u = \sqrt{\frac{3gh}{2} }

t = h / u

= \sqrt{\frac{2h}{3g} }

Distance covered by A

= 1/2 gt²

= 1/2 x 2h/3

= h/3

Height at which collision occured

= h - h/3

= 2h /3

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salantis [7]

Answer:

9 cm

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Explanation:

u = Object distance

v = Image distance

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Lens equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm

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v=-4\times 9=-36\ cm

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3 years ago
Fill in a T/F answer for each statement below. If false, correct the statement to make it true.:
grandymaker [24]

Answer:

Explanation:

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3 0
3 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

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