consider the mirror from the last question. an object 4cm tall stands 10cm in front of a converging mirror of focal length 5cm.
2 answers:
Answer:
This question is incomplete, here is the complete question;
"Consider the mirror shown here. An object 4 cm tall stands 10 cm in front of a converging mirror of focal length 5 cm.
The image formed by this mirror is: (Select all that apply.)
real
virtual
enlarged
the same size
diminished
Inverted
erect"
Answer: Virtual, Upright and Enlarged
Explanation:
The formula for image formed by a converging mirror is given by;
------------------- eqn 1
where f = focal length = 5 cm
u = object height = 4 cm
v = image height = ?
substitute for u and f in eqn 1 to find v.
v = -20 cm
From the value of the image formed i.e -20 cm
The negative sign means the image is behind the mirror and this means image is;
1. Virtual
2. Upright
3. Enlarged
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momentum(p) = mass(m) x velocity(v)
KE = kinetic energy = 1/2 mv²
a.
the ratio of the momentum of the SUV to that of the car
b. the ratio of the KE of the SUV to that of the car
Answer:
Explanation:
a ) If the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) , the image from object lens must have been formed at the focus of eye lens . So the objective image must have been formed at 19.5 - 2.75 = 16.75 cm from the object lens.
b ) Let the object distance be u
For object lens
v = 16.75 cm , f = .35 cm
1/v - 1/u = 1/f
1/16.75 - 1/u = 1/ .35
.0597 - 1/u = 2.857
1/u = - 2.7973
u = .3575 cm
c ) Angular magnification
=
v₀ and u₀ are image and object distance for object lens , D = 25 cm and f_e is focal length of eye lens
= (16.75 / .3575) x( 25 / 2.75)
= 46.85 x 9.09
= 426