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Oliga [24]
2 years ago
7

Two arrows are shot vertically upward. the second arrow is shot after the first one, but while the first is still on its way up.

the initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. suppose that the initial speed of the first arrow is 31.8 m/s and that the second arrow is fired 1.82 s after the first. determine the initial speed of the second arrow.
Physics
1 answer:
Neko [114]2 years ago
7 0
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get....... 17.15 m/s</span>
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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

4 0
2 years ago
A trumpet makes sound when the lips of the musician vibrate. True False
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Block A weighs 14 lb and block B weighs 5 lb If B is moving downward with a velocity (vB)1 = 3 ft/s at t = 0, determine the velo
kolezko [41]

Velocity, va2 = 10.5 ft/s

<u>Explanation:</u>

From the figure:

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∴ vₐ = -2vb

Applying the principle of Impulse and momentum in x-direction

mv_x_1 + \int\limits^t_t {F_x} \, dt = mv_x_2

Limit is t1 to t2

-(\frac{14}{32.2}) (2) (3) - T = \frac{14}{32.2} v_y_2                           -(1)

Applying the principle of Impulse and momentum in y-direction

mv_y_1 + \int\limits^t_t {F_y} \, dt = mv_y_2

Limit is t1 to t2

(\frac{5}{32.2}) (3) - 2T + 3 = -\frac{5}{32.2} \frac{v_y_2}{2}                            -(2)

Solving equation (1) and (2), we obtain

T = 1.6lb

va2 = 10.5 ft/s

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