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klemol [59]
3 years ago
11

I have no idea of how to approach this problem

Physics
1 answer:
nataly862011 [7]3 years ago
4 0

Answer:

p=1

Explanation:

Well me know that v=m/s

and that a=m/s^2

so

(m/s)^{2} = (m/s^2)(x^p)\\\\(m^2/s^2)/(m/s^2)=x^p\\\\m^2s^2/ms^2=x^p\\\\(m^2/m)(s^2/s^2)=x^p\\\\m=x^p\\\\p=1

Note: We don't take into account 2 because it's a scalar, it doesn't have units so it doesn't add anything to the equation.

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We can calculate the density of the balloon as follows:

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Therefore, the balloon will fall

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In which point the electric intensity of a sphere is maximum​
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What is the mechanical advantage of a 8 m ramp that rises 2 m to a stage?
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Given the following data;

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Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole,
tamaranim1 [39]

Answer:

The ratio of E_{app} and E_{act} is 0.9754

Explanation:

Given that,

Distance z = 4.50 d

First equation is

E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}

E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}

Second equation is

E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}

We need to calculate the ratio of E_{act} and E_{app}

Using formula

\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}

\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}

Put the value into the formula

\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}

\dfrac{E_{app}}{E_{act}}=0.9754

Hence, The ratio of E_{app} and E_{act} is 0.9754

8 0
3 years ago
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