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Zolol [24]
3 years ago
7

If you push a 100-kilogram box on Earth, then take it to the Moon and try to push it there, it would feel like you were pushing

a _________-kilogram box
a. 16
b. 100
c. 160
d. 600
Physics
2 answers:
Vesna [10]3 years ago
8 0
A. i believe because the gravitational pull is stronger on earth than it is on the moon therefore making the box lighter than its original weight.
Sunny_sXe [5.5K]3 years ago
5 0
It would feel like your puching a 16 kilogram box
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The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A
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D. 2^(3/2)

Explanation:

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T(Y)² = x³

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The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

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Uranium

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Two waves travel through ocean, one with wavelength 3 m, one with 6 m. Which one has greater speed?
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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
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