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4 years ago

Explain why significant figures represent the precision of a measurement and not its accuracy

1 answer:

3 0

Firstly, when we say precision, this means the level of detail of a number. On the other hand, significance means that how precise a number is written and accuracy is defined as how close the number is to an actual value. Significant figures represent the precision of a measurement and not its accuracy because when a measurement has a larger number of significant figures, it has a greater reproducibility <span>or precision because it has a smaller source of error in the estimated digit. Hope this answer helps.</span>

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A thin, uniformly charged insulating rod has a linear charge density λ = 3 nC/m and lies along the x axis from x = 1m to x = 3m.

Contact [7]

Answer:

A) $V_A = 11.93~V$

B) The vector definition of E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

where magnitude is E = 2.66 N/m.

Explanation:

The potential of a uniformly charged rod can be found by the method of integration. We will first choose an infinitesimal part on the rod. We will compute the potential of this part at point A. Then we will integrate this potential over the entire rod.

We will use the following formula for electric potential:

$V = \frac{1}{4\pi \epsilon_0}\frac{Q}{r}$

Let us choose the infinitesimal part a distance 'x' from the origin. Then the distance between this point and point A is

$r = \sqrt{x^2+4^2}$

The infinitesimal length is 'dx', and the potential of this length is dV. Let's apply the formula:

$dV = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{\sqrt{x^2 + 4^2}}$

Here, the charge Q is equal to the charge density multiplied by the length. Q = λdx

Now we have to integrate this infinitesimal potential over the rod:

$V = \int\limits^3_1 {dV} \, dx = \frac{1}{4\pi \epsilon_0}\int\limits^3_1 {\frac{\lambda}{\sqrt{x^2 + 16}} \, dx$

By using an integral table, this can be calculated:

$V = \frac{3\times 10^{-9}}{4\pi\epsilon_0}\ln(|\sqrt{x^2+16}+x|)\left \{ {{x=3} \atop {x=1}} \right. \\V = 11.93~V$

B) The electric field can be found by a similar approach, but a different formula:

$\vec{E} = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}\^r$

Let's apply this formula to the infinitesimal part we have chosen.

$dE_x = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\cos(\theta)\\dE_y = \frac{1}{4\pi\epsilon_0}\frac{\lambda dx}{x^2 + 4^2}\sin(\theta)$

By the geometry sine and cosine terms can be found:

$\sin(\theta) = \frac{4}{\sqrt{x^2+16}}\\\cos(\theta) = \frac{x}{\sqrt{x^2 + 16}}$

The x- and y-components of the E-field can be found separately by integrating the infinitesimal parts over the entire rod.

$E_x = \int\limits^3_1 {dE_x} \, dx = \frac{\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{x}{(x^2+16)^{3/2}}} \, dx = 1.13(-\^x)\\E_y = \int\limits^3_1 {dE_y} \, dx = \frac{4\lambda}{4\pi\epsilon_0}\int\limits^3_1 {\frac{1}{(x^2+16)^{3/2}}} \, dx = 2.41(\^y)$

So, the final E-field is

$\vec{E} = -1.13\^x + 2.41\^y$

The magnitude of the E-field is

E = 2.66 N/m

6 0

3 years ago

PLS HELPPPP AHHHH I WILL MARK AS BRAINLIEST

SIZIF [17.4K]

Answer:

Liquid

Explanation:

To turn liquid into gas, you add heat, to remove heat from gas, you make liquid.

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2 years ago

One star might appear brighter than another star because

Marrrta [24]

Answer:

A star's brightness also depends on its proximity to us. The more distant an object is, the dimmer it appears. Therefore, if two stars have the same level of brightness, but one is farther away, the closer star will appear brighter than the more distant star - even though they are equally bright!

Can I have brainliest?

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3 years ago

Two people, one with mass m1 and the other with mass m2, stand on a stationary sled with mass M on a frozen lake. Assume that th

lozanna [386]

Answer:

Part a)

Velocity of sled

$v = \frac{m_1 s}{m_1 + m_2 + M}$