4. A 905 kg car travels around a circular track with a circumference of 3.25 km. If the magnitude of the centripetal force is 2140 N, what is the car's tangential speed?

To find the car's tangential speed, we can use the formula for centripetal force:<br /><br />\[ F_c = \frac{mv^2}{r} \]<br /><br />where:<br />- \( F_c \) is the centripetal force,<br />- \( m \) is the mass of the car,<br />- \( v \) is the tangential speed, and<br />- \( r \) is the radius of the circular track.<br /><br />Given:<br />- \( F_c = 2140 \) N,<br />- \( m = 905 \) kg, and<br />- \( r = \frac{3.25 \text{ km}}{2\pi} \) (since the circumference is \( 2\pi r \)).<br /><br />First, let's calculate the radius \( r \):<br /><br />\[ r = \frac{3250 \text{ m}}{2\pi} \approx 517.7 \text{ m} \]<br /><br />Now, we can rearrange the centripetal force formula to solve for the tangential speed \( v \):<br /><br />\[ v = \sqrt{\frac{F_c \cdot r}{m}} \]<br /><br />Plugging in the values:<br /><br />\[ v = \sqrt{\frac{2140 \text{ N} \cdot 517.7 \text{ m}}{905 \text{ kg}}} \]<br /><br />\[ v \approx \sqrt{\frac{111,148.8 \text{ Nm}}{905 \text{ kg}}} \]<br /><br />\[ v \approx \sqrt{122.9 \text{ m}^2/\text{s}^2} \]<br /><br />\[ v \approx 11.1 \text{ m/s} \]<br /><br />Therefore, the car's tangential speed is approximately 11.1 m/s.