Answer:
distance between object and image = 18.9 cm
Explanation:
given data
radius of curvature = 18 cm
focal length = 1/2 radius of curvature
magnification = 40%
to find out
distance between object and image
solution
we know lens formula that is
1/f = 1/v + 1/u ....................1
here f = 18 /2 and v and u is object and image distance
and we know m = 40% = 0.40
so 0.40 = -v / u
so here v = - 0.40 u
so from equation 1
1/f = 1/v + 1/u
2/18 = - 1/0.40u + 1/u
u = -13.5 cm ..................2
and
v = -0.40 (- 13.5)
v = 5.4 cm ......................3
so from equation 2 and 3
distance between object and image = 5.4 + 13.5
distance between object and image = 18.9 cm
You first subtract the speed at which the man is moving (11 m/s) from the rate the boat is moving (12.4 m/s). Which equals 1.4, then divide it by 6 meters, as the man is moving relative to the boat.
It therefore equals 4.29 s
Let v = the running speed
After running at constant speed for 26 min, the distance traveled is
d = (v m/min)*(26 min) = 26v m
Because there are 1500 m to go, the distance traveled is
10000 - 1500 = 8500 m
The running speed is
v = (8500 m)/(26 min) = 326.9 m/min
In km/h, the speed is
v = (0.3269 km/min)*(60 min/h) = 19.6 km/h
Answer: The running speed is 19.6 km/h
Answer:
In the Equator:
As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.
In Antarctic:
As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.
Explanation: