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3 years ago

12

We have seen that an electric field must exist inside a conductor that carries a current. How is that possible in view of the fa

ct that in electrostatics we concluded that the electric field must be zero inside a conductor?

1 answer:

sleet_krkn [62]3 years ago

7 0

Answer:

Explanation:

The potential difference between one side of the wire causes the electric field inside the wire (causes the electrons to flow). However, inside the wire, it is still neutral. The electrons are just moving, the wire is not gaining or losing electrons.

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False. They are arranged in a structure called a crystal lattice

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Where does energy originate ina food web

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A rifle fires a bullet at a velocity of 78 m/s, 40 degrees above the horizontal. Determine the height (h) above the starting pos

qwelly [4]

Answer:

H = Vy t - 1/2 g t^2 height of an object with an initial "vertical" velocity

at t sec after firing

Vy = 78 m/s * sin 40 = .643 * 78 m/s = 50.1 m/s

H = 50.1 * 6 - 1/2 * 9.8 * 6^2 = 300 m - 176 m = 124 m

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3 years ago

A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o

Margarita [4]

Answer:

V_{average} = $\frac{1}{2} V_o$ , V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

V (t) = $\left \{ {{V=V_o \ \ \ t< \tau /2} \atop {V=0 \ \ \ \ t> \tau /2 } } \right.$

to substitute in the equation let us separate the into two pairs

V_average = $\int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt$

V_average = $V_o \ \int\limits^{1/2}_0 {} \, dt$

V_{average} = $\frac{1}{2} V_o$

we evaluate V₀ = 4 V

V_{average} = 4 / 2)

V_{average} = 2 V

6 0

2 years ago

A ball is thrown directly downward with an initial speed of 7.70 m/s, from a height of 30.2 m. After what time interval does it

KatRina [158]

Answer:

$t = 1.82$

Explanation:

Given

$u = 7.70m/s$ -- initial velocity

$s = 30.2m$ --- height

Required

Determine the time to hit the ground

This will be solved using the following motion equation.

$s = ut + \frac{1}{2}gt^2$

Where

$g = 9,8m/s^2$

So, we have:

$30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2$

$30.2 = 7.70t + 4.9 * t^2$

Subtract 30.2 from both sides

$30.2 -30.2 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9 * t^2 - 30.2$

$0 = 7.70t + 4.9t^2 - 30.2$

$7.70t + 4.9t^2 - 30.2 = 0$

$4.9t^2 + 7.70t - 30.2 = 0$

Solve using quadratic formula:

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 4.9;\ b = 7.70;\ c = -30.2$

$t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}$

$t = \frac{-7.70\±\sqrt{651.21}}{9.8}$

$t = \frac{-7.70\±25.52}{9.8}$

Split the expression

$t = \frac{-7.70+25.52}{9.8}$ or $t = \frac{-7.70-25.52}{9.8}$

$t = \frac{17.82}{9.8}$ or $t = -\frac{33.22}{9.8}$

Time can't be negative; So, we have:

$t = \frac{17.82}{9.8}$

$t = 1.82$

Hence, the time to hit the ground is 1.82 seconds

7 0

2 years ago