Answer:

Explanation:
given data:
refractive index of lens 1.50
focal length in air is 30 cm
focal length in water is -188 cm
Focal length of lens is given as
![\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%5Cfrac%7Bn_2%20-n_1%7D%7Bn_1%7D%20%2A%20%5Cleft%20%5B%5Cfrac%7B1%7D%7Br1%7D%20-%5Cfrac%7B1%7D%7Br2%7D%20%20%20%5Cright%20%5D)
![\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%5Cfrac%7Bn_%7Bg%7D%20-n_%7Bair%7D%7D%7Bn_%7Bair%7D%7D%20%2A%20%5Cleft%20%5B%5Cfrac%7B1%7D%7Br1%7D%20-%5Cfrac%7B1%7D%7Br2%7D%20%20%20%5Cright%20%5D)
![\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%5Cfrac%7Bn_%7Bg%7D%20-1%7D%7B1%7D%20%2A%20%5Cleft%20%5B%5Cfrac%7B1%7D%7Br1%7D%20-%5Cfrac%7B1%7D%7Br2%7D%20%20%20%5Cright%20%5D)
focal length of lens in liquid is
![\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%5Cfrac%7Bn_%7Bg%7D%20-n_%7Bl%7D%7D%7Bn_%7Bl%7D%7D%20%2A%20%5Cleft%20%5B%5Cfrac%7B1%7D%7Br1%7D%20-%5Cfrac%7B1%7D%7Br2%7D%20%20%20%5Cright%20%5D)

rearrange fro



Answer:

Explanation:
As we know that the equation of particle position is given as

Now the speed of the particle is given as

now we know that potential energy and kinetic energy is equal
so we have

so we will have






Answer:
the magnitude of the displacement after 5s is 137.31 m.
Explanation:
Given;
initial velocity of the projectile, u = 60 m/s
angle of projection, θ = 60°
time of motion, t = 5s
the vertical component of the velocity, 
The magnitude of the displacement after 5s is calculated as;

Therefore, the magnitude of the displacement after 5s is 137.31 m.
Answer: the answer is 100 focal length.
Explanation: mark brainliest PLEASE