What is a mirror's radius of curvature if cars 17.0 m away appear 0.35 their normal size? follow the sign conventions.

The Mirror's radius of curvature is 18.45 m

Explanation:

Radius of curvature (R)  = 2 x Focal length (F)

Focal length = R / 2

Object distance (u) = 17 m

magnification = image distance (v) / Object distance (u)

                0.35 = v / u

                  v = 0.35 u

Mirror formula

1 / F = 1 / u + 1 / v

1 /F = 1 /17 + 1 / -0.35u

1/F  = 1/17 + 1  / -5.98

1/F  = 1 /17 - 1 /5.98

1 / F = - 0.1084

F = - 9.225 m

Focal length is negative because the image formed is virtual.

Recall that: Radius of curvature (R)  = 2 x Focal length (F)

                                                            = 2 x 9.225

                     Radius of curvature (R) = 18.45 m

pe = weightxheight = 2000x1000=2,000,000joules

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures