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Tomtit [17]
3 years ago
7

You walk 10 miles north, 20 miles south, and 5 miles north. Taking north to be the positive y direction, define a position vecto

r that describes your position on the y-axis.
Physics
1 answer:
gulaghasi [49]3 years ago
3 0

Considering north as positive Y axis

       First  10 miles north means a positive 10 displacement along y direction

      Second 20 miles south means a negative 20 displacement along y direction

       Third 5 miles south means a positive 5 displacement along y direction

So total displacement = 10 - 20 + 5 = -5 miles displacement

So position vector of the final position = - 5 j

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Full and New Moon

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How many joules of kinetic energy does a pendulum have when it has 100 joules of potential energy
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The maximum kinetic energy is 100 j.    

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<h3>The kinetic energy = (potential energy) + (kinetic energy) and the potential energy of 0 J implying its kinetic energy is 100 J, which is its maximum. </h3>

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A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the r
Murrr4er [49]

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2.1 rad/s

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Mass of a tether ball, m = 0.546 kg

Length of a rope, l =  4.56 m

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F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s

Let \omega is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :

v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s

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4 0
3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
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