Why are units of measurement useful?

2 answers:

5 0

Without the ability to measure, it would be difficult for scientists to conduct experiments or form theories. Not only is measurement important in science and the chemical industry, it is also essential in farming, engineering, construction, manufacturing, commerce, and numerous other occupations and activities.

kondor19780726 [428]4 years ago

4 0

They set a common stadard by which all involved can understand the context of what is being evaluated.

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In a rectangular coordinate system a positive point charge q = 6.00 x 10^-9C is placed at the point x = +0.150 m, y = 0, and an

makvit [3.9K]

Answer:

a) 0N/C

b) 2660.7N/C

c) 543.7N/C

Explanation:

To find the electric field in all these cases you take into account the x and y component of the total Electric field for each point.

for P(0,0), only there is the x component of the field because the point is the parallel line that connects both charges:

$E_x=E_{1x}-E_{2x}\\\\E_x=k\frac{q}{r_1^2}-k\frac{q}{r_2^2}$ ( 1 )

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r1: distance to the first charge = 0.150m

r:2 distance to the second charge = 0.150m

Due to in this case the distance r1=r2: you obtain, by replacing in (1):

$E_x=0$

for P(0.300 , 0 ) you also have only the x component of E, and the electric field generated by each charge are directed toward right:

$\vec{E}=E_x\hat{i}$

$E_x=k\frac{q}{r_1^2}+k\frac{q}{r_2^2}$

r1 = 0.300m+0.150m = 0.450m

r2 = 0.300m-0.150m = 0.150m

By replacing r1 and r2 in ( 1 ) you obtain:

$E_{x}=kq(\frac{1}{r_1^2}+\frac{1}{r_2^2})=(8.98*10^9Nm^2/C^2)(6.00*10^{-9})(\frac{1}{(0.450m)^2}+\frac{1}{(0.150m)^2})\\\\E_x=2660.7N/C\\\\\vec{E}=2660.7N/C\ \hat{i}$

for P(0.150 , -0.40) you have both x and y components for E:

$\vec{E}=E_x\hat{i}+E_y\hat{j}\\\\E_x=k\frac{q}{r_1^2}cos\theta+0N/C\\\\E_y=-k\frac{q}{r_1^2}sin\theta-k\frac{q}{r_2^2}$

the second charge does not contribute for the x component of E.

To find r1 you use Pitagora's theorem:

$r_1=\sqrt{(0.150+0.150m)^2+(0.40m)^2}=0.500m$

r2 = 0.40m

the angle is obtain by using a simple trigonometric relation:

$tan\theta=\frac{0.40}{0.150}=2.66\\\\\theta=tan^{-1}(2.66)=69.44\°$

Then, by replacing the values of r1, r1, q, theta and k you obtain:

$E_x=(8.98*10^9Nm^2/C^2)\frac{(6.00*10^{-9}C)}{(0.500m)^2}cos69.44=75.68N/C\\\\E_y=-(8.98*10^9Nm^2/C^2)(6.00*10^{-9}C)(\frac{sin69.44}{(0.500m)^2}+\frac{1}{(0.40m)^2})\\\\E_y=538.42N/C\\\\\vec{E}=75.68N/C\hat{i}-538.42N/C\hat{j}\\\\|\vec{E}|=\sqrt{(E_x)^2+(E_y)^2}=543.7N/C\\\\\theta=tan^{-1}(\frac{538.42}{75.68})=278\°$