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NikAS [45]
3 years ago
10

Me Ayudan con este ejercicio por favor !!!

Physics
1 answer:
klasskru [66]3 years ago
6 0
Tom llega primero a la superficie de el lago
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Two rams run toward each other. One ram has a mass of 44 kg and runs south with a speed of 6 m/s, while the other has a mass of
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A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo
guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}

If we consider than the direction of the electric field is \vec{E}=E_0\hat{x}, we can solve the problem differentiating  the integral for each face of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6

E₀ is a constant and each surface is equal to each other, so: S_1=S_2=S_i=S

Therefore:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c

3 0
3 years ago
An all-electric car (not a hybrid) is designed to run from a bank of 12.0 V batteries with total energy storage of 2.30 ✕ 107 J.
AnnyKZ [126]

Answer:

a) I=733.33\ A

b) d=52272.7273\ m

c) d'=51948.0519\ m

Explanation:

Given:

  • voltage of the battery, V=12\ V
  • energy storage capacity of the battery, E=2.3\times 10^7\ J
  • speed of the car, v=20\ m.s^{-1}

a)

power drawn by the car, P=8.8\ kW

<u>Now the Current delivered to the motor:</u>

we the relation between the power and electrical current,

P=V.I

8800=12\times I

I=733.33\ A

b)

<u>Distance travelled before battery is out of juice:</u>

we first find the time before the battery runs out,

t=\frac{E}{P}

t=\frac{2.3\times 10^7}{8800}

t=2613.636\ s

Now the distance:

d=v.t

d=20\times 2613.636

d=52272.7273\ m

c)

When the head light of 55 W power is kept on while moving then the power   consumption of the car is:

P'=P+55

P'=8800+55

P'=8855\ W

<u>Now the time of operation of the car:</u>

t'=\frac{E}{P'}

t'=\frac{2.3\times 10^7}{8855}

t'=2597.4026\ s

<u>Now the distance travelled:</u>

d'=v.t'

d'=20\times 2597.4025

d'=51948.0519\ m

5 0
3 years ago
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