At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m. A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted. C) Determine the magnification of the image.

Explanation:

For a convex mirror, the value of its image distance and its focal length are negative.

using the mirror formula 1/f = 1/u+1/v

f is the focal length = Radius of curvature/2 = 0.560/2

f= 0.28m

u is the object distance = 10.6m

v is the position of the image = ?

On substitution;

1/0.28 = 1/10.6 + 1/-v

3.57 = 0.094 - 1/v

3.57 - 0.094 = -1/v

3.476 = -1/v

v = -1/3.476

v = -0.2877m

B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.

C) Magnification = Image distance/object distance

Magnification  = 0.2877/10.6

Magnification = 0.0271

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j

the awnser is d you right bro