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insens350 [35]
2 years ago
14

You toss a walnut at a speed of 15.0 m/s at an angle of 50.0∘ above the horizontal. The launch point is on the roof of a buildin

g that is 20.0 m above the ground.
a)How long after it is launched does the walnut reach the ground?

b)How far does the walnut travel horizontally from launch point to landing point?

c) What are the horizontal and vertical components of the walnut’s velocity just before it reaches the ground? Suppose that the +y-axis is directed upward and the +x-axis is directed horizontally from the launch point toward the landing point.
Physics
1 answer:
kaheart [24]2 years ago
8 0

Answer:

a.3.51s

b.33.8m

c. 9.64,-22.9

Explanation:

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A cork shoots out of a champagne bottle at an angle of 40.0 above the horizontal. If the cork travels a horizontal distance of 1
ruslelena [56]

Answer:

The initial speed of the cork was 1.57 m/s.

Explanation:

Hi there!

The equation of the horizontal position of the cork in function of time is the following:

x = x0 + v0 · t · cos θ

Where:

x = horizontal position at time t.

x0 = initial horizontal position.

v0 = initial speed of the cork.

t = time.

θ = launching angle.

If we place the origin of the frame of reference at the launching point, then x0 = 0.

We know that at t = 1.25 s, x = 1.50 m. We also know the launching angle so we can solve the equation of horizontal position for the initial speed, v0:

x = v0 · t · cos θ

x / t · cos θ = v0

v0 = 1.50 m / (1.25 s · cos (40.0°)

v0 = 1.57 m/s

The initial speed of the cork was 1.57 m/s.

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3 years ago
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What unit is used for the value of G in Newton's Law of Universal<br> Gravitation?<br> *
Helga [31]

Answer:

In SI units, its value is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The modern notation of Newton's law involving G was introduced in the 1890s by C. V. Boys. The first implicit measurement with an accuracy within about 1% is attributed to Henry Cavendish in a 1798 experiment.

Explanation:

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Two scientists are discussing their beliefs about something they cannot observe
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Could it be hypothesis?
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If a bar of copper is brought near a magnet, the copper bar will be
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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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