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A police officer in hot pursuit drives her car through a circular turn of radius 379 mwith a constant speed of 90.0 km/h. Her mass is 52.0 kg.What are (a) themagnitude and (b) the angle (relative to vertical) of the net force of the officer onthe car seat? (Hint: Consider both horizontal and vertical forces.)

A Police Officer In Hot Pursuit Drives Her Car Through A Circular Turn Of Radius 379 Mwith A Constant Speed Of 900 Kmh Her Mass Is 520 KgWhat Are A Themagnitude class=

Sagot :

The centripetal force keeps the car in the circle. This force is directed towards the center of the circle. The upward force on the police officer is f = mg and the centripetal force is represented as follows

[tex]F_c=\frac{mv^2}{r}[/tex]

The net force will be

[tex]\begin{gathered} F_{net}=F_c+mg \\ F_{net}=\sqrt[]{(\frac{mv^2}{r})^2+(mg)^2} \end{gathered}[/tex]

A.

[tex]\begin{gathered} r=379m \\ m=52.0\text{ kg} \\ \text{speed}=90\text{ km/h} \\ \text{speed}=25\text{ m/s} \\ F_{net}=\sqrt[]{(\frac{52\times25^2}{379})^2+(52\times9.8)^2} \\ F_{net}=\sqrt[]{(\frac{32500}{379})^2+(509.6)^2} \\ F_{net}=\sqrt[]{(85.7519788918)^2+259692.16} \\ F_{net}=\sqrt[]{7353.40188386+259692.16} \\ F_{net}=\sqrt[]{267045.561884} \\ F_{net}=516.764512988 \\ F_{net}\approx516.76N \end{gathered}[/tex]

B.

[tex]\begin{gathered} \tan \emptyset=\frac{\frac{mv^2}{r}}{mg} \\ \tan \emptyset=\frac{mv^2}{r}\times\frac{1}{mg} \\ \tan \emptyset=\frac{v^2}{rg} \\ \tan \emptyset=\frac{25^2}{379\times9.8} \\ \tan \emptyset=\frac{625}{3714.2} \\ \tan \emptyset=0.16827311399 \\ \emptyset=\tan ^{-1}0.16827311399 \\ \emptyset=9.55185382438 \\ \emptyset=9.55^{\circ} \end{gathered}[/tex]

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