A 4.0-cm tall light bulb is placed at distance of 8.3 cm from a concave mirror having a focal length of 15.2 cm. Determine the i
mage distance and the image size. Also indicate the orientation of the image compared to the object as well as if it will be a real or virtual image.
1 answer:
Answer: the image distance is -18, 28 cm this means behind of the concave mirror. The image size is 2.2 higher that the original so it has 8.8 cm with the same orientation as original and it is a virtual imagen.
Explanation: In order to sove the imagen formation for a concave mirror we have to use the following equation:
1/p+1/q=1/f where p and q represents the distance to the mirror for the object and imagen, respectively. f is the focal length for the concave mirror.
replacing the values we obtain:
1/8.3+1/q=1/15.2
so 1/q=(1/15.2)-(1/8.3)=-54.7*10^-3
then q=-18.28 cm
The magnification is given by M=-q/p=-(-18,28)/8.3= 2.2
We also add a picture to see the imagen formation for this case.
You might be interested in
Answer:
The magnitude of the force F is given by
F = ( +
+
) *(
*g/(
))
Explanation:
Given there are three blocks of masses ,
and
(ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( +
+
) *a ................(equation 1)
Also it is given that does not move with respect to
, which gives tension T is exerted on pulley by
only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T =
................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
*
=
* (
+
)
*
= (
*
)+ (
*
)
( -
) *
=
*
=
*
/(
-
)
Taking square root on both sides, we get acceleration a
a = *g/(
)
Hence substituting the value of a in equation 1, we get
F = ( +
+
) *(
*g/(
))
