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Alekssandra [29.7K]
3 years ago
13

si se deja caer un carrito desde el punto mas alto de ua psta de coches cuya altura es de 1.4m cual es la velocidad maxima que p

uede alcanzar el carrito?
Physics
1 answer:
forsale [732]3 years ago
6 0

Answer:

v = 5.24[m/s]

Explanation:

Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

E_{kin}=E_{pot}

Donde:

E_{kin}=\frac{1}{2} *m*v^{2} \\\\E_{pot}=m*g*h

Ahora reemplazando:

\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} }   \\\\v=5.24[m/s]

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Two small metal spheres are 25 cm apaft.The spheres have equal amount of negative charge and repel each other with a force of 0.
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Answer:

0.5\times 10^{-6}C

Explanation:

According to coulombs law force between two charges is given by F=\frac{1}{4\pi \epsilon _0}\frac{Q_1Q_2}{R^2} here R is the distance between both the charges which is given as 25 cm

We have given force F =0.036 N

So  0.036=\frac{1}{4\pi \times 8.85\times 10^{-12}}\frac{Q^2}{(0.25^2)} As \epsilon _0 is constant which value is 8.85\times 10^{-12}

Q^2=0.250\times 10^{-12}

Q=0.5\times 10^{-6}C

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3 years ago
The electric field in the region between the plates of a parallel plate capacitor has a magnitude of 6.3 105 V/m. If the plate s
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Answer:

3.528×10² V.

Explanation:

potential difference: This is the work done when one coulomb of charge moves from one point to another in an electric field. The S.I unit of potential difference is volt. The formula of potential difference is given as,

V = kq/r..................... Equation 1

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Comparing equation 1 and equation 2,

V = E×r............................. Equation 3

Where V = potential difference, E = Electric field between the plate of the capacitor, r = distance between the plate.

Given: E = 6.3×10⁵ V/m, r = 0.56 mm = 0.00056 m.

Substitute into equation 3,

V = 6.3×10⁵×0.00056

V = 3.528×10² V.

Hence the potential difference of the plate = 3.528×10² V.

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