Answer:
The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.
Explanation:
Considering west direction along negative x-axis and north direction along positive y-axis
Given:
The car travels at a speed of 120 km/h in the west direction.
The car then travels at the same speed in the north direction.
Now, considering the given directions, the velocities are given as:
Velocity in west direction is, ![\overrightarrow{v_1}=-120\ \vec{i}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv_1%7D%3D-120%5C%20%5Cvec%7Bi%7D)
Velocity in north direction is, ![\overrightarrow{v_2}=120\ \vec{j}](https://tex.z-dn.net/?f=%5Coverrightarrow%7Bv_2%7D%3D120%5C%20%5Cvec%7Bj%7D)
Now, since
are perpendicular to each other, their resultant magnitude is given as:
![|\overrightarrow{v_{res}}|=\sqrt{|\overrightarrow{v_1}|^2+|\overrightarrow{v_2}|^2}](https://tex.z-dn.net/?f=%7C%5Coverrightarrow%7Bv_%7Bres%7D%7D%7C%3D%5Csqrt%7B%7C%5Coverrightarrow%7Bv_1%7D%7C%5E2%2B%7C%5Coverrightarrow%7Bv_2%7D%7C%5E2%7D)
Plug in the given values and solve for the magnitude of the resultant.This gives,
![|\overrightarrow{v_{res}}|=\sqrt{(120)^2+(120)^2}\\\\|\overrightarrow{v_{res}}|=120\sqrt{2} = 169.71\ km/h](https://tex.z-dn.net/?f=%7C%5Coverrightarrow%7Bv_%7Bres%7D%7D%7C%3D%5Csqrt%7B%28120%29%5E2%2B%28120%29%5E2%7D%5C%5C%5C%5C%7C%5Coverrightarrow%7Bv_%7Bres%7D%7D%7C%3D120%5Csqrt%7B2%7D%20%3D%20169.71%5C%20km%2Fh)
Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.
So, the direction is given as:
![x=\tan^{-1}(\frac{|v_2|}{|v_1|})\\\\x=\tan^{-1}(\frac{120}{-120})=\tan^{-1}(-1)=-45\ deg(clockwise\ angle\ with\ the\ x-axis)](https://tex.z-dn.net/?f=x%3D%5Ctan%5E%7B-1%7D%28%5Cfrac%7B%7Cv_2%7C%7D%7B%7Cv_1%7C%7D%29%5C%5C%5C%5Cx%3D%5Ctan%5E%7B-1%7D%28%5Cfrac%7B120%7D%7B-120%7D%29%3D%5Ctan%5E%7B-1%7D%28-1%29%3D-45%5C%20deg%28clockwise%5C%20angle%5C%20with%5C%20the%5C%20x-axis%29)
Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.
i do not have an answer because it depends on the size and the distance lol
Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is
![\sqrt{3} :\sqrt{10}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%3A%5Csqrt%7B10%7D)
Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e
![R\propto l](https://tex.z-dn.net/?f=R%5Cpropto%20l)
- inversely proportional to its cross section area i.e
![R\propto \frac{1}{A}](https://tex.z-dn.net/?f=R%5Cpropto%20%5Cfrac%7B1%7D%7BA%7D)
Therefore
![R=\rho\frac{l}{A}](https://tex.z-dn.net/?f=R%3D%5Crho%5Cfrac%7Bl%7D%7BA%7D)
ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:
![A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r_1%5E2%20%3D%5Cpi%20%28%5Cfrac%7Bd_1%7D%7B2%7D%20%29%5E2)
![R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }](https://tex.z-dn.net/?f=R_1%3D%5Crho_1%5Cfrac%7Bl_1%7D%7B%5Cpi%28%5Cfrac%7Bd_1%7D%7B2%7D%29%5E2%20%7D)
......(1)
Again for tungsten:
![R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }](https://tex.z-dn.net/?f=R_2%3D%5Crho_2%5Cfrac%7Bl_2%7D%7B%5Cpi%28%5Cfrac%7Bd_2%7D%7B2%7D%29%5E2%20%7D)
........(2)
Given that
and ![l_1=l_2](https://tex.z-dn.net/?f=l_1%3Dl_2)
Dividing the equation (1) and (2)
![\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}](https://tex.z-dn.net/?f=%5CRightarrow%5Cfrac%7B%20%28%5Cfrac%7Bd_1%7D%7B2%7D%29%5E2%7D%7B%20%28%5Cfrac%7Bd_2%7D%7B2%7D%29%5E2%7D%3D%5Cfrac%7B%5Crho_1%5Cfrac%7Bl_1%7D%7B%5Cpi%20R_1%20%7D%7D%7B%5Crho_2%5Cfrac%7Bl_2%7D%7B%5Cpi%20R_2%20%7D%7D)
[since
and
]
![\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}](https://tex.z-dn.net/?f=%5CRightarrow%28%20%5Cfrac%7Bd_1%7D%7Bd_2%7D%20%29%3D%5Csqrt%7B%5Cfrac%7B1.68%5Ctimes%2010%5E%7B-8%7D%7D%7B5.6%5Ctimes%2010%5E%7B-8%7D%7D%7D)
![\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}](https://tex.z-dn.net/?f=%5CRightarrow%28%20%5Cfrac%7Bd_1%7D%7Bd_2%7D%20%29%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B10%7D%7D)
![\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}](https://tex.z-dn.net/?f=%5CRightarrow%20d_1%3Ad_2%3D%5Csqrt%7B3%7D%20%3A%5Csqrt%7B10%7D)
Therefore the ratio of diameter of the copper to that of the tungsten is
![\sqrt{3} :\sqrt{10}](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20%3A%5Csqrt%7B10%7D)
P = U × I
I = P / U = 1600W / 120V = 13.4A
P = 240V × 13.4A = 3216W
If your hair blower isnt rated for 220- 230V(this is the voltage in EU) you are most likely going to burn it.
Answer:
Explanation:
They need a galvanic difference. Or saying that less technically, they need to have different electron attraction, so that one can collect electrons (oxidation/reduction) and flow current from the other. :)