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Contact [7]
2 years ago
8

15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers

are listed in the image. Thank you so much!

Physics
1 answer:
timofeeve [1]2 years ago
3 0

The percent difference between two numbers x and y is given by

\dfrac{|y-x|}x \times 100\%

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from x to y or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

\dfrac{|6-3|}3 \times 100\% = 100\%

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

\dfrac{3-6}3\times100\%=-50\%

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take x from the left column and y from the right column.

\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%

\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%

\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%

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Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, d=1.9\times 10^7\ m

Mass of Earth, M=5.97\times 10^{24}\ kg

Radius of Earth, r=6.37\times 10^6\ m

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Drag the correct labels to the box. Not all labels will be used. Identify the examples of persuasive writing. We must act to pre
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–0.05 m/s

Explanation:

The total momentum of the system player+basketball must be conserved before and after the ball has been thrown.

Before throwing the ball, the total momentum of the system is zero, because can assume both the player and the basketball being at rest:

p_i

The total momentum after the ball has been thrown is instead the sum of the momenta of the the player and of the basketball:

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m_p = 59 kg is the player's mass

v_p is the player's velocity

m_b=0.51 kg is the ball's mass

v_b=6 m/s is the ball's velocity

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0=m_p v_p + m_b v_b

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