Answer:
d_{b} = 2 d_{a}
Explanation:
The electrical resistance for a cylindrical wire is described by the expression
R = ρ L / A
The area of a circle is
A = π r²
r = d / 2
A = π d²/4
We substitute
R = ρ L 4 /π d²
Let's apply this expression to our case, they indicate that the resistance of wire A is 4 times the resistance of wire B
= 4 R_{b}
We substitute
ρ 4/π
² = 4 (ρ 4/π d_{b}²)
1 / d_{a}² = 4 / d_{b}²
d_{a} = d_{b} / 2
Answer: Lever
A wheelbarrow consists of a lever to be able to lift the material, and a wheel to be able to move it horizontally. So in a sense, a wheelbarrow is a complex machine consisting of two simple machines.
A shadow forms on the side of an object that faces away from the sun. The length of shadows changes as Earth rotates. In the morning, the sun is low in the eastern sky and shadows are long. As time passes in the morning, the sun seems to move higher in the sky.
Answer:
(a) The horizontal ground reaction force ![F_{g,x}=325\, N](https://tex.z-dn.net/?f=F_%7Bg%2Cx%7D%3D325%5C%2C%20N)
(b) The vertical ground reaction force ![F_{g,y}=696\, N](https://tex.z-dn.net/?f=F_%7Bg%2Cy%7D%3D696%5C%2C%20N)
(c) The resultant ground reaction force ![F_g=768\, N](https://tex.z-dn.net/?f=F_g%3D768%5C%2C%20N)
Explanation:
Given
John mass , m = 65 kg
Horizontal acceleration , ![a_x= 5.0 \frac{m}{s^{2}}](https://tex.z-dn.net/?f=a_x%3D%205.0%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
Vertical acceleration , ![a_y=0.9 \frac{m}{s^{2}}](https://tex.z-dn.net/?f=a_y%3D0.9%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D)
(a) Using Newton's 2nd law in horizontal direction
![F_{g,x}=ma_x](https://tex.z-dn.net/?f=F_%7Bg%2Cx%7D%3Dma_x)
=>![F_{g,x}=65\times 5\, N=325\, N](https://tex.z-dn.net/?f=F_%7Bg%2Cx%7D%3D65%5Ctimes%205%5C%2C%20N%3D325%5C%2C%20N)
Thus the horizontal ground reaction force ![F_{g,x}=325\, N](https://tex.z-dn.net/?f=F_%7Bg%2Cx%7D%3D325%5C%2C%20N)
(b) Using Newton's 2nd law in vertical direction
![F_{g,y}-mg=ma_y](https://tex.z-dn.net/?f=F_%7Bg%2Cy%7D-mg%3Dma_y)
=>![F_{g,y}=mg+ma_y](https://tex.z-dn.net/?f=F_%7Bg%2Cy%7D%3Dmg%2Bma_y)
=>![F_{g,y}=65\times (9.81+0.9)\, N=696\, N](https://tex.z-dn.net/?f=F_%7Bg%2Cy%7D%3D65%5Ctimes%20%289.81%2B0.9%29%5C%2C%20N%3D696%5C%2C%20N)
Thus the vertical ground reaction force ![F_{g,y}=696\, N](https://tex.z-dn.net/?f=F_%7Bg%2Cy%7D%3D696%5C%2C%20N)
(c) Resultant ground reaction force is
![F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}](https://tex.z-dn.net/?f=F_g%3D%28F_%7Bg%2Cx%7D%5E%7B2%7D%2BF_%7Bg%2Cy%7D%5E%7B2%7D%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D)
=>![F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N](https://tex.z-dn.net/?f=F_g%3D%28325%5E%7B2%7D%2B696%5E%7B2%7D%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%5C%2C%20N%3D768%5C%2C%20N)
=>![F_g=768\, N](https://tex.z-dn.net/?f=F_g%3D768%5C%2C%20N)
Thus the resultant ground reaction force ![F_g=768\, N](https://tex.z-dn.net/?f=F_g%3D768%5C%2C%20N)