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## Solution:

The weight of an object is computed using the formula $\displaystyle \text{Weight}=\text{mass}\times 9.81\:\text{m/s}^2$

### Part A $\displaystyle \text{Weight}=8\:\text{kg}\times 9.81\:\text{m/s}^2$ $\displaystyle \text{Weight}=78.5\:\text{newtons}$          ◀

### Part B $\displaystyle \text{Weight}=0.04\:\text{kg}\times 9.81\:\text{m/s}^2$ $\displaystyle \text{Weight}=0.3924\:\text{newtons}$          ◀

### Part C $\displaystyle \text{Weight}=760\:\text{Mg}\times \frac{1000\:\text{kg}}{1\:\text{Mg}}\times 9.81\:\text{m/s}^2$ $\displaystyle \text{Weight}=7\:455\:600\:\text{newtons}$          ◀

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## College Physics 3.67 – The angle of a puck’s velocity relative to the player

#### An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the player is 29.0 m/s. The line between the center of the goal and the player makes a 90.0º angle relative to his path as shown in Figure 3.63. What angle must the puck’s velocity make relative to the player (in his frame of reference) to hit the center of the goal? Figure 3.63 An ice hockey player moving across the rink must shoot backward to give the puck a velocity toward the goal.

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