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
Answer:
25 seconds
Explanation:
Assuming the woman is accelerating at a constant rate of
from the initial velocity, u=4.20 m/s, to the final velocity, v=5.00 m/s.
Let she takes t seconds to cover the distance, s=115 m.
As acceleration, ![a=\frac{v-u}{t}=\frac{5-4.2}{t}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bv-u%7D%7Bt%7D%3D%5Cfrac%7B5-4.2%7D%7Bt%7D)
![\Rightarrow at=0.8\cdots(i)](https://tex.z-dn.net/?f=%5CRightarrow%20at%3D0.8%5Ccdots%28i%29)
Now, from the equation of motion
![s=ut+\frac 12 at^2](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%2012%20at%5E2)
![\Rightarrow s=ut+\frac 12 at(t)](https://tex.z-dn.net/?f=%5CRightarrow%20s%3Dut%2B%5Cfrac%2012%20at%28t%29)
[ from equation (i)]
![\Rightarrow 115=(4.2+0.4)t](https://tex.z-dn.net/?f=%5CRightarrow%20115%3D%284.2%2B0.4%29t)
seconds.
Hence, she takes 25 seconds to walk the distance.
I believe the answer is c
Answer:
Hope this helps you find the answer
Explanation:
The proteins, lipids, and polysaccharides that make up most of the food we eat must be broken down into smaller molecules before our cells can use them—either as a source of energy or as building blocks for other molecules.