<span>1. sandpaper against wood.
2. rubber mat against the floor of the shower stall.
</span><span># High coefficient of friction describes the degree of interaction between two surfaces. . A higher coefficient of friction indicates that two surfaces in contact have a greater resistance.</span>
Complete Question:
Find the resistance of a wire of length 0.65 m, radius 0.25 mm and resistivity 3 * 10^{-6} ohm-metre.
Answer:
Resistance = 9.95 Ohms
Explanation:
<u>Given the following data;</u>
Length = 0.65 m
Radius = 0.25 mm to meters = 0.00025 m
Resistivity = 3 * 10^{-6} ohm-metre.
To find the resistance of the wire;
Mathematically, resistance is given by the formula;

Where;
- P is the resistivity of the material.
- L is the length of the material.
- A is the cross-sectional area of the material.
First of all, we would find the cross-sectional area of the wire.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.00025)²
Area = 3.142 * 6.25 * 10^{-8}
Area = 1.96 * 10^{-7} m²
Now, to find the resistance of the wire;


<em>Resistance = 9.95 Ohms </em>
Answer:
a)
, b)
, c) 
Explanation:
a) The change in the gravitational potential energy of the marble-Earth system is:


b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

c) The spring constant of the gun is:




Answer:
I think it would be a chemical change.
Explanation:
Because a physical would be like crushing it or something along that line.
Answer: 170.67 N
Explanation:
Given
Mass of skier is 
Height of the inclination is 
Here, the potential energy of the skier is converted into kinetic energy which is consumed by the friction force by applying a constant force that does work to stop the skier.
![\Rightarrow mgh=F\cdot x\quad \quad [\text{F=constant friction force}]\\\\\Rightarrow 82.9\times 9.8\times 20=F\cdot 95.2\\\\\Rightarrow F=\dfrac{16,248.4}{95.2}\\\\\Rightarrow F=170.67\ N](https://tex.z-dn.net/?f=%5CRightarrow%20mgh%3DF%5Ccdot%20x%5Cquad%20%5Cquad%20%5B%5Ctext%7BF%3Dconstant%20friction%20force%7D%5D%5C%5C%5C%5C%5CRightarrow%2082.9%5Ctimes%209.8%5Ctimes%2020%3DF%5Ccdot%2095.2%5C%5C%5C%5C%5CRightarrow%20F%3D%5Cdfrac%7B16%2C248.4%7D%7B95.2%7D%5C%5C%5C%5C%5CRightarrow%20F%3D170.67%5C%20N)
Thus, the horizontal friction force is 170.67 N.