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4 years ago

9

How to find the gradient of a velocity time graph

1 answer:

Sidana [21]4 years ago

7 0

The area-

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

<span><span>Area of light-blue triangle -
<span>The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: <span>area of triangle = 1⁄2 × base × height </span><span>so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. </span></span></span><span> Area of dark-blue rectangle The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m.</span><span> Area under the whole graph <span>The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.<span>This is the total area under the distance-time graph. This area represents the distance covered.</span></span></span></span>

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sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

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Charge of particle 2 = $1.50\times10^{-7} c$

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Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

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