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dimulka [17.4K]
3 years ago
8

The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared ​(units are in​ meters). If

v Subscript x is constant at 5.0 ​m/s, find the resultant velocity at the point left parenthesis 2.0 comma 3.0 right parenthesis .
Physics
1 answer:
ivanzaharov [21]3 years ago
8 0

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

y=3+0.8x-0.4x^2........(1)

The x component of constant velocity, v_x=5\ m/s

We need to find the resultant velocity at the point (2,3).

Let \dfrac{dx}{dt}=v_x and \dfrac{dy}{dt}=v_y

Differentiating equation (1) wrt t as,

\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}

v_y=0.8\times v_x-0.8x\times v_x

v_y=0.8v_x(1-x)

When x = 2 and v_x=5\ m/s

So,

v_y=0.8\times 5\times (1-2)

v_y=-4\ m/s

Resultant velocity, v=\sqrt{v_x^2+v_y^2}

v=\sqrt{5^2+(-4)^2}

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

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Explanation:

Given:

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<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

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Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

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<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

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E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

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E_y=0\ N.C^{-1}

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74.86°C

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