The ratio of the distance moved by the point at which the effort is applied in a simple machine to the distance moved by the point at which the load is applied, in the same time. In the case of an ideal (frictionless and weightless) machine, velocity ratio = mechanical advantage. Velocity ratio is sometimes called distance ratio.
Answer:
a
The focal length of the lens in water is 
b
The focal length of the mirror in water is 
Explanation:
From the question we are told that
The index of refraction of the lens material = 
The index of refraction of the medium surrounding the lens = 
The lens maker's formula is mathematically represented as
![\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%28n%20-1%29%20%5B%5Cfrac%7B1%7D%7BR_1%7D%20-%20%5Cfrac%7B1%7D%7BR_2%7D%20%20%5D)
Where
is the focal length
is the index of refraction
are the radius of curvature of sphere 1 and 2 of the lens
From the question When the lens in air we have
![\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7Bair%7D%7D%20%3D%20%28n-1%29%20%5B%5Cfrac%7B1%7D%7BR_1%7D%20-%20%5Cfrac%7B1%7D%7BR_2%7D%20%20%5D)
When immersed in liquid the formula becomes
![\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf_%7Bwater%7D%7D%20%3D%20%5B%5Cfrac%7Bn_2%7D%7Bn_1%7D%20-%201%20%5D%20%5B%5Cfrac%7B1%7D%7BR_1%7D%20-%20%5Cfrac%7B1%7D%7BR_2%7D%20%20%5D)
The ratio of the focal length of the the two medium is mathematically evaluated as
![\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }](https://tex.z-dn.net/?f=%5Cfrac%7Bf_water%7D%7Bf_%7Bair%7D%7D%20%3D%20%5Cfrac%7Bn_2%20-1%7D%7B%5B%5Cfrac%7Bn_2%7D%7Bn_1%7D%20-%201%5D%20%7D)
From the question
= 79.0 cm

and the refractive index of water(material surrounding the lens) has a constant value of 


b
The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.
Answer:
12.0 meters
Explanation:
Given:
v₀ = 0 m/s
a₁ = 0.281 m/s²
t₁ = 5.44 s
a₂ = 1.43 m/s²
t₂ = 2.42 s
Find: x
First, find the velocity reached at the end of the first acceleration.
v = at + v₀
v = (0.281 m/s²) (5.44 s) + 0 m/s
v = 1.53 m/s
Next, find the position reached at the end of the first acceleration.
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²
x = 4.16 m
Finally, find the position reached at the end of the second acceleration.
x = x₀ + v₀ t + ½ at²
x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²
x = 12.0 m
Answer: 
Explanation:
The acceleration of an object can be calculated by using Newton's second law:

where
F is the net force applied on the object
m is the mass of the object
a is its acceleration
In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:

This is too late too answer?