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Marianna [84]
4 years ago
9

The image shows the path of a ball from the time it’s thrown to the time it lands on the ground. Determine the kind of energy th

e ball has at each position. (PE stands for gravitational potential energy, and KE stands for kinetic energy.)
PE
neither
PE and KE

Physics
1 answer:
LenaWriter [7]4 years ago
4 0

At Position 1: The ball has both PE and KE. (The ball is at a height from ground and velocity is given to the ball)

At Position 2: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.

At Position 3: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 4: The ball has only PE. (The ball is at a height from ground and velocity is zero.)

At position 5: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 6: The ball has both PE and KE. (The ball is at a height from ground and velocity is also there.)

At position 7: The ball has only KE (Just before hitting the ground). After landing the ball have neither of the energy.

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Explanation:

(a) The minimum value of x will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

h^{2} = b^{2} + p^{2}

h^{2} = 5^{2} + 0.17^{2}   \\h = \sqrt{} 25.03\\h= 5.002 m

<em>Hence, the maximum distance is 5.002 m</em>

(c) For minimum magnitude we use the minimum distance calculated in (a)

Minimum Distance = 0.17 m

For electrostatic force=     F=\frac{kq1q2}{x^{2} }

F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19}  }{0.17^{2} }

F= 6.38×10^{-26} N

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F =  \frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19}   }{5.002^{2} } }

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Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

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mass of the 70 ft rope  = 2 lb/ft x 70 ft

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