Answer:
Where is question 12, we need it to answer this question
Explanation:
Answer:
q = -2 m and q = -0.5 m
Explanation:
For this exercise we must use the equation of the optical constructor
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distance to the object and the image, respectively
Let's start with the far vision point, in this case the power of the lens is
P = -0.5D
power is defined as the inverse of the focal length in meter
f = 1 / D
f = -1 / 0.5
f = - 2m
the object for the far vision point is at infinity p = infinity
1 / f = 1 / p + i / q
1 / q = 1 / f - 1 / p
1 / q = -1/2 - 1 / ∞
q = -2 m
The sign indicates that the image is on the same side as the object
Now let's lock the near view point
D = +2.00 D
f = 1 / D
f = 0.5m
the near mink point is p = 25 cm = 0.25 m
1 / f = 1 / p + 1 / q
1 / q = 1 / f - 1 / p
1 / q = 1 / 0.5 - 1 / 0.25
1 / q = -2
q = -0.5 m
the sign indicates that the image is on the same side as the object in front of the lens
If a ball is if a ball is dropped from a 576ft building it would take about 8 seconds for it to hit the ground.
Answer:
work done = force x distance
Explanation:
F = 23 N
D = 2m
W = 23 * 2 = 46 J
<u>Answer:</u> The Young's modulus for the wire is 
<u>Explanation:</u>
Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.
The equation representing Young's Modulus is:
where,
Y = Young's Modulus
F = force exerted by the weight = 
m = mass of the ball = 10 kg
g = acceleration due to gravity = 
l = length of wire = 2.6 m
A = area of cross section = 
r = radius of the wire = 
(Conversion factor: 1 m = 1000 mm)
= change in length = 1.99 mm = 
Putting values in above equation, we get:
Hence, the Young's modulus for the wire is
