Explanation:
Given that,
Object-to-image distance d= 71 cm
Image distance = 26 cm
We need to calculate the object distance


We need to calculate the focal length
Using formula of lens

put the value into the formula



The focal length of the lens is 35.52.
(B). Given that,
Object distance = 95 cm
Focal length = 29 cm
We need to calculate the distance of the image
Using formula of lens

Put the value in to the formula




We need to calculate the magnification
Using formula of magnification



The magnification is 0.233.
The image is virtual.
Hence, This is the required solution.
<span>Here I think you have to find the velocity in x and y components where x is east and y is north
So as air speed indicator shows the negative speed in y component and adding it in
air speed while multiplying with the direction component we will get the velocity as velocity is a vector quantity so direction is also required
v=-28 m/s y + 18 m/s (- x/sqrt(2) - y/sqrt(2))
solving
v= -12.7 m/s x-40.7 m/s y
if magnitude of velocity or speed is required then
speed= sqrt(12.7^2 + 40.7^2)
speed= 42.63 m/s
if angle is asked
angle = arctan (40.7/12.7)
angle = 72.67 degrees south of west</span>
Answer:
1.23 m
Explanation:
The vertical distance covered by a free-falling object starting from rest in a time t is

where
g = 9.8 m/s^2 is the acceleration due to gravity
In this problem, we have
t = 0.50 s
So the distance covered is
