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dalvyx [7]
3 years ago
11

An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

R away from the wire using Gauss's Law for Electrostatics.
Physics
1 answer:
marshall27 [118]3 years ago
4 0

Answer:

E = \frac{\lambda}{2\pi \epsilon_0 r}

Explanation:

Let the linear charge density of the charged wire is given as

\frac{q}{L} = \lambda

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

\int E. dA = \frac{q}{\epsilon_0}

here we have

E \int dA = \frac{\lambda L}{\epsilon_0}

E. 2\pi r L = \frac{\lambda L}{\epsilon_0}

so we have

E = \frac{\lambda}{2\pi \epsilon_0 r}

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Licemer1 [7]

Answer:

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Explanation:

Because you would need to be able to handle the weight to move it.

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4 years ago
A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration
JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on  and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m

So, the maximum height reached by the rocket is:

h=y_1+y_2\\h=160m+154.28m=314.28m

(c) In the first part we have:

v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s

And in the second part:

t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s

So,  the time it takes to reach the maximum height is:

t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}

t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s

So, the total time is:

t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s

7 0
4 years ago
Whoever gets this right gets brainlyess answer and it's worth 10 points :)
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A is only the area of one tile.  It wants the area of the whole floor.
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Which type of climate receives the most intense solar energy? A.
Natali5045456 [20]
I think D is correct
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4 years ago
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