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3 years ago

7

A proton is released from rest inside a region of constant, uniform electric field E1 pointing due North. 32.3 seconds after it

is released, the electric field instantaneously changes to a constant, uniform electric field E2 pointing due South. 3.45 seconds after the field changes, the proton has returned to its starting point. What is the ratio of the magnitude of E2 to the magnitude of E1? You may neglect the effects of gravity on the proton.

1 answer:

NARA [144]3 years ago

8 0

Answer:

E₂ / E₁ = 521.64 / 5.95 =87.67

Explanation:

Let d be the distance covered inside electric field . Lt q be the magnitude of charge.

Force under field E₁ = q E₁

acceleration = qE₁/ m

d = 1/2 a t²

d = .5 ( qE₁ / m) x 32.3²

d = 521.64 ( qE₁ / m)

Similarly for return journey,

d = .5 x ( qE₂ / m) x 3.45²

d = 5.95x( qE₂ / m)

521.64 ( qE₁ / m) = 5.95x( qE₂ / m)

E₂ / E₁ = 521.64 / 5.95 =87.67

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Si un planeta tuviese un periodo de traslación de 65 años terrestres a que distancia se encontraría del sol

Tju [1.3M]

Answer:

$R \approx 2.418\times 10^{9}\,km$

Explanation:

(The following exercise is written in Spanish and for that reason explanation will be held in Spanish)

Supóngase que el planeta tiene una órbita circular, el período de rotación del planeta es:

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Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

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Ahora se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }$

La aceleración experimentada por el planeta es:

$a_{r} = G\cdot \frac{M_{sun}}{R^{2}}$

Se reemplaza en la ecuación de período:

$T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }$

La distancia del planeta con respecto al sol es finalmente despejada:

$R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}$

$R = \sqrt[3]{G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}}$

Finalmente, se sustituyen las variables y se determina la distancia:

$R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}$

$R \approx 2.418\times 10^{12}\,m$

$R \approx 2.418\times 10^{9}\,km$

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3 years ago

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Answer:

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Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which the speed of the boat in km / h? And in m / s?

Given that,

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or

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So, current drawn is given as:

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So, current drawn is given as:

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