Ways to increase friction
- increase the area of contact
- increase the roughness of the contact materials
- dry up or take away any lubricant
- increase the pressure on the contact
Answer with explanation:
The Normalization Principle states that

Given
Thus solving the integral we get

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

Applying the limits and solving for A we get
![I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B1%7D%7Bk%7D%5B%5Cfrac%7B1%7D%7Be%5E%7Bkx%7D%7D-%5Cfrac%7Bx%7D%7Be%5E%7Bkx%7D%7D%5D_%7B0%7D%5E%7B%2B%5Cinfty%20%7D%5C%5C%5C%5CI%3D-%5Cfrac%7B1%7D%7Bk%7D%5C%5C%5C%5C%5Ctherefore%20A%3D-k)
Answer:
11.76Newtons
Explanation:
Workdone = Force * distance
Given
Workdone = 180joules
Distance = 15.3 metres
Required
Force
From the formula;
Force - Work/distance
Force = 180/15.3
Force = 11.76Newtons
Hence the required force is 11.76Newtons
Explanation:
Let thermal capacity of the vessel be C' J K-1
Heat energy given by hot water = 40 x 4.2 x (60 - 30) = 5040 J
Heat energy taken by cold water = 50 x 4.2 x (30 - 20) = 2100 J
Heat energy taken by vessel = C' x (30 - 20) = 10 C' J
If there is no loss of heat energy,
Heat energy given by hot water = Heat energy taken by cold water and vessel
or 5040 = 2100 + 10 C'
or 10 C' = 2940
or C'= 294 J K-1
Thus, thermal capacity of vessel = 294 J K-1.
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Answer:
i wanna say tranverse if not surface
Explanation: