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mash [69]
4 years ago
7

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the s

ky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1061 N. The mass of the sky diver is 93.4 kg. Take upward to be the positive direction. What is his acceleration, including sign
Physics
1 answer:
ser-zykov [4K]4 years ago
6 0

Explanation:

According to newton's second law of motion.

\sum Fx = ma\\\\\sum Fx = 1061 - 915\\\\\sum Fx = 146N

m is the mas of the sky diver = 93.4kg

a is the acceleration of the skydiver

From the formula above;

a = \frac{\sum Fx}{m}\\ \\a = \frac{146}{93.4}\\\\a = 1.563m/s^2

Hence the acceleration of the sky diver is 1.563m/s²

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Answer:

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\overrightarrow{v}=\overrightarrow{v_{1}}+\overrightarrow{v_{2}}

\overrightarrow{v}=\left (31.7+42.26  \right )\widehat{i}+\left ( 67.97- 90.63 \right )\widehat{j}

\overrightarrow{v}=73.96\widehat{i}-22.66\widehat{j}

magnitude of resultant velocity is \sqrt{\left ( 73.96 \right )^{2}+\left ( -22.66 \right )^{2}}

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