You will use the Pythagorean Theorem to solve it.
c^2 = a^2 + b^2
c^2 = (1.5)^2 + (2)^2
c^2 = 6.25
c = square root of 6.25
c = 2.5
I hope this helps!
Answer: f=150cm in water and f=60cm in air.
Explanation: Focal length is a measurement of how strong light is converged or diverged by a system. To find the variable, it can be used the formula:
= (nglass - ni)(
-
).
nglass is the index of refraction of the glass;
ni is the index of refraction of the medium you want, water in this case;
R1 is the curvature through which light enters the lens;
R2 is the curvature of the surface which it exits the lens;
Substituting and calculating for water (nwater = 1.3):
= (1.5 - 1.3)(
-
)
= 0.2(
)
f =
= 150
For air (nair = 1):
= (1.5 - 1)(
-
)
f =
= 60
In water, the focal length of the lens is f = 150cm.
In air, f = 60cm.
Answer: 2.1 × 10^7 m/s
Explanation:
Please see the attachments below
Answer:
The workdone is 
Explanation:
From the question we are told that
The height of the cylinder is 
The face Area is 
The density of the cylinder is 
Where
is the density of freshwater which has a constant value

Now
Let the final height of the device under the water be 
Let the initial volume underwater be 
Let the initial height under water be 
Let the final volume under water be 
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as


So 
=> 
Now the work done is mathematically represented as

![= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.](https://tex.z-dn.net/?f=%3D%20%20%20%5Crho_w%20g%20A%20%5B%5Cfrac%7Bh%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20h_f%7D%20%5Catop%20%7Bh%7D%7D%20%5Cright.)
![= \frac{g A \rho}{2} [h^2 - h_f^2]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%20%5Bh%5E2%20-%20h_f%5E2%5D)
![= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%28h%5E2%29%20%20%5B1%20%20-%20%5Cfrac%7Bh_f%5E2%7D%7Bh%5E2%7D%20%5D)
Substituting values
