clicking submit, you are agreeing to the Allen High School Honor Code s' statement. TheAllen High School Honor Code statement reads "On my honor, s en High School student,I have neither given nor recelved unauthorized assistance: Question 4 If you have a 0.98 kg ball traveling in a circle with radius 2.99 m ata velocity of 2.16 m/s.how much Is the centripetal force on the ball? square 10 pts

To find the centripetal force on the ball, we can use the formula:<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 ball,<br />- \( v \) is the velocity of the ball,<br />- \( r \) is the radius of the circle.<br /><br />Given:<br />- \( m = 0.98 \, \text{kg} \)<br />- \( v = 2.16 \, \text{m/s} \)<br />- \( r = 2.99 \, \text{m} \)<br /><br />Now, plug these values into the formula:<br /><br />\[ F_c = \frac{(0.98 \, \text{kg}) \times (2.16 \, \text{m/s})^2}{2.99 \, \text{m}} \]<br /><br />First, calculate the numerator:<br /><br />\[ (0.98 \, \text{kg}) \times (2.16 \, \text{m/s})^2 = 0.98 \times 4.6656 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 4.556288 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \]<br /><br />Next, divide by the radius:<br /><br />\[ F_c = \frac{4.556288 \, \text{kg} \cdot \text{m}^2/\text{s}^2}{2.99 \, \text{m}} = 1.523 \, \text{N} \]<br /><br />Therefore, the centripetal force on the ball is approximately \( 1.523 \, \text{N} \).