Friction can be defined as a type of force that prevents two smooth surfaces from sliding together on each other. Friction is a strong force that oposses motion when two surfaces come in contact with each other.
The different methods/ways to reduce friction include;
- Application of lubricant to surfaces
- Streamlined bodies; this goes a long way in preventing friction between two bodies.
- Decrease in weight; reducing the amount of weight on a substance can also help to reduce friction
- Applying wheels to objects that move/roll on the ground.
The magnitude of normal force that floor exerts on chair is 100.83N
<u>Explanation:</u>
Given:
Weight of chair, W = 75N
Applied force, F = 42N
Angle, θ = 38°
Normal force, n = ?
We know,
Vertical component of the force = F sinθ
= 42 X sin38°
= 42 X 0.615
= 25.83N
Total normal force acting on the chair = 75N + 25.83N
= 100.83N
Therefore, the magnitude of normal force that floor exerts on chair is 100.83N
Answer:
The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.
Explanation:
Given that,
Wavelength = 617 nm
Width of slit ![d= 6.30\times10^{-6}\ m](https://tex.z-dn.net/?f=d%3D%206.30%5Ctimes10%5E%7B-6%7D%5C%20m)
Distance between the slit and screen L= 2.83 m
Third dark fringe m = 3
We need to calculate the distance on the screen from the center of the central bright fringe to the third dark fringe on either side
Using formula of distance
![d \sin\theta=m\lambda](https://tex.z-dn.net/?f=d%20%5Csin%5Ctheta%3Dm%5Clambda)
![\sin\theta=\dfrac{y}{l}](https://tex.z-dn.net/?f=%5Csin%5Ctheta%3D%5Cdfrac%7By%7D%7Bl%7D)
Put the value into the formula
![d\times\dfrac{y}{L}=m\times\lambda](https://tex.z-dn.net/?f=d%5Ctimes%5Cdfrac%7By%7D%7BL%7D%3Dm%5Ctimes%5Clambda)
![y=\dfrac{m\times\lambda\timesL}{d}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7Bm%5Ctimes%5Clambda%5CtimesL%7D%7Bd%7D)
![y=\dfrac{3\times617\times10^{-9}\times2.83}{6.30\times10^{-6}}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B3%5Ctimes617%5Ctimes10%5E%7B-9%7D%5Ctimes2.83%7D%7B6.30%5Ctimes10%5E%7B-6%7D%7D)
![y=0.831\ m](https://tex.z-dn.net/?f=y%3D0.831%5C%20m)
Hence, The distance on the screen from the center of the central bright fringe to the third dark fringe is 0.831 m.