Aconcave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) find the location of the image for object distances of (i) 40.0 cm, (ii) 20.0 cm, and (iii) 10.0 cm. for each case, state whether the image is (b) real or virtual and (c) upright or inverted. (d) find the magnification in each case.

Explanation:

It is given that,

Radius of curvature of the mirror, R = 20 cm

So, focal length of the mirror, f = -10 cm

(i) Object distance, u = -40 cm

Using the mirror's formula as :

v is the image distance

v = -13.33 cm

Magnification of mirror is calculated as :

m = -0.33

Since, the magnification is negative, image is real and inverted.

(ii) Object distance, u = -20 cm

Using the mirror's formula as :

v is the image distance

v = -20 cm

Magnification of mirror is calculated as :

m = -0.5

Since, the magnification is negative, image is real and inverted.

(iii) Object distance, u = -10 cm

Using the mirror's formula as :

v is the image distance

v = infinity

Magnification of mirror is calculated as :

magnification = infinity

Hence, this is the required solution.

the chart show the masses and velocities of two colliding objects that stick together after a collision.

according to the law of conservation of momentum, the momentum of the object after the collision is  

answer

1,500

kg • m/s.

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//