Answer:
The value is ![p = - 2.63 \ Diopters](https://tex.z-dn.net/?f=p%20%3D%20%20%20-%202.63%20%5C%20Diopters)
Explanation:
From the question we are told that
The value of the far point is ![a = 40 \ cm = 0.4 \ m](https://tex.z-dn.net/?f=a%20%3D%20%2040%20%5C%20cm%20%20%3D%20%200.4%20%5C%20%20m)
The distance of the lens to the eye is ![b = 2 \ cm = 0.02](https://tex.z-dn.net/?f=b%20%3D%20%202%20%5C%20cm%20%3D%200.02)
Generally
![1 Diopter = > 1 m^{-1}](https://tex.z-dn.net/?f=1%20Diopter%20%3D%20%3E%20%201%20m%5E%7B-1%7D)
Generally the power spectacle lens needed is mathematically represented as
![p = \frac{1}{d_o } + \frac{1}{d_i}](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B1%7D%7Bd_o%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7Bd_i%7D)
Here
is the object distance which for a near sighted person is
And
is the image distance which is evaluated as
![d_i = b - a](https://tex.z-dn.net/?f=d_i%20%3D%20%20b%20-%20a)
=> ![d_i = 0.02 - 0.4](https://tex.z-dn.net/?f=d_i%20%3D%20%200.02%20-%200.4)
=> ![d_i = -0.38 \ m](https://tex.z-dn.net/?f=d_i%20%3D%20-0.38%20%5C%20%20m)
So
![p = \frac{1}{\infty } + \frac{1}{-0.38}](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B1%7D%7B%5Cinfty%20%7D%20%20%2B%20%5Cfrac%7B1%7D%7B-0.38%7D)
=> ![p = 0 - 2.63](https://tex.z-dn.net/?f=p%20%3D%200%20%20%20-%202.63)
=> ![p = - 2.63 \ Diopters](https://tex.z-dn.net/?f=p%20%3D%20%20%20-%202.63%20%5C%20Diopters)
Answer:
A. 1.71 m
B. 2.66 m
Explanation:
A. Determination of the height of the pier.
We'll begin by calculating the time taken for the ball to get to the water
This can be obtained as follow:
Horizontal velocity (u) = 1.27 m/s,
Horizontal distance (s) = 0.75 m
Time (t) =?
s = ut
0.75 = 1.27 × t
Divide both side by 1.27
t = 0.75 / 1.27
t = 0.59 s
Finally, we shall determine the height of the pier as follow:
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 0.59 s
Height of pier (h) =?
h = ½gt²
h = ½ × 9.8 × 0.59²
h = 4.9 × 0.3481
h = 1.71 m
Thus, the height of the pier is 1.71 m
B. Determination of the horizontal distance.
Horizontal velocity (u) = 4.50 m/s
Time (t) = 0.59 s
Horizontal distance (s) =?
s = ut
s = 4.5 × 0.59
s = 2.66 m
Thus, if the ball moved at a velocity of 4.50 m/s off the pier, it will land at a distance of 2.66 m from the end of the pier.
Answer:
X
Explanation:
v=u+at but t is constant so
a=v-u
for x
a=29-10
a=19
which marks the greater acceleration
Answer:
A sonar device sends pulses of sound waves down through the water. When these pulses hit objects like fish, vegetation or the bottom, they are reflected back to the surface. The sonar device measures how long it takes for the sound wave to travel down, hit an object and then bounce back up.