First find Acceleration
- Initial velocity=u=0m/s
- Final velocity=v=42m/s
- Time=t=2s
- Distance=s
- Acceleration=a
![\boxed{\sf Acceleration=\dfrac{v-u}{t}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20Acceleration%3D%5Cdfrac%7Bv-u%7D%7Bt%7D%7D)
![\\ \sf\longmapsto Acceleration=\dfrac{42-0}{2}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D%5Cdfrac%7B42-0%7D%7B2%7D)
![\\ \sf\longmapsto Acceleration=\dfrac{42}{2}](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D%5Cdfrac%7B42%7D%7B2%7D)
![\\ \sf\longmapsto Acceleration=21m/s^2](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20Acceleration%3D21m%2Fs%5E2)
Using second equation of kinematics
![\boxed{\sf s=ut+\dfrac{1}{2}at^2}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2%7D)
![\\ \sf\longmapsto s=0(2)+\dfrac{1}{2}(21)(2)^2](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20s%3D0%282%29%2B%5Cdfrac%7B1%7D%7B2%7D%2821%29%282%29%5E2)
![\\ \sf\longmapsto s=21(2)](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20s%3D21%282%29)
![\\ \sf\longmapsto s=42m](https://tex.z-dn.net/?f=%5C%5C%20%5Csf%5Clongmapsto%20s%3D42m)
The size of the field 'decreases'. Imagine zooming in on your TV so that your face is pressed up against the screen. you can see the individual pixels but you cannot see the whole picture because the sizer of your 'field of view' has shrunk.
Answer:
Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A
Explanation:
Let the depth of each section is h.
That means the real depth for each section is h.
Apparent depth is liquid A is 7 cm.
Apparent depth in liquid B is 6 cm.
Apparent depth in liquid C is 5 cm.
by the formula of the refractive index
n = real depth / apparent depth
where, n is the refractive index of the liquid.
For liquid A:
.... (1)
For liquid B:
..... (2)
For liquid C:
..... (3)
By comparing all the three equations
nc > nB > nA
Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A
Answer:
Distance, S = 440 meters.
Explanation:
Given the following data;
Initial velocity, u = 17m/s
Time, t = 20 seconds
Final velocity, v = 27m/s
To find the distance;
First of all, we would determine the acceleration of the truck.
Acceleration = (v-u)/t
Substituting the given values into the equation, we have;
Acceleration = (27 - 17)/20
Acceleration = 10/20
Acceleration = 0.5m/s²
Now, we would use the second equation of motion to find the distance traveled.
S = ut + ½at²
S = 17*20 + ½*0.5*20²
S = 340 + 0.25*400
S = 340 + 100
S = 440m