An object that is 2.00 cm high is placed 10.0 cm in front of a concave mirror with a radius of curvature of 40.0 cm. Find the magnification and location of the corresponding image in relation to the mirror’s surface.

The magnification is 1, that is the image is upright and the same size as the object

The location of the image is behind the mirror (virtual image)

Explanation:

Here we have the mirror equation presented as follows;

Where:

= Distance of the object from the mirror = 10.0 cm

= Distance of the image from the mirror

R = Radius of curvature of the mirror = 40.0 cm

f = Focal length of the mirror = 1/2 × R = 1/2 × 40 = 20 cm

m = Magnification of the mirror

= Height of the object = 2.00 cm

= Height of the image

Plugging in the values, we have

∴ = -10 cm which means a virtual image, behind the mirror

The magnification is found and presented as follows;

Therefore, the image is upright and the same size as the object.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.