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daser333
1 month ago
14

An infinite charged wire with charge per unit length lambda lies along the central axis of a cylindrical surface of radius r and

length . What is the flux through the surface due to the electric field of the charged wire?
Physics
1 answer:
inna [3.1K]1 month ago
3 0
The electric flux through the cylindrical surface surrounding the infinite charged wire is given by the formula ∅E = E x 2πrl. To analyze this, we consider an infinitely long straight wire with a uniform linear charge density of λ Cm⁻¹. The electric field at a distance r from this charge can be evaluated using a cylindrical Gaussian surface of radius r and length l, oriented along the wire. Only the curved surface of the cylinder contributes to the total flux since the other surfaces are perpendicular to E.
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An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z
Sav [3153]

Refer to the attached file for the solution

7 0
3 months ago
For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo
ValentinkaMS [3465]
The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height. 

<span>In this scenario, you can demonstrate it as follows: </span>

<span>the initial velocity is Vo </span>
<span>the launch angle is α </span>

<span>the initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity becomes </span>
<span>Vh = Vo×cos(α) </span>

<span>the total flight duration is the period required to return to a height of 0 m, thus </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time =? </span>
<span>a = gravitational acceleration = g (= -9.8 m/s²) </span>
<span>therefore </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now let's examine the horizontal distance. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range =? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>therefore </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
4 0
1 month ago
A person who climbs up something (e.g., a hill, a ladder, the stairs) from the ground gains potential energy. a person's weight
kicyunya [3294]

The following values have been provided:

weight w = 240 lb = 1,067.52 N

energy E = 3,000 J

 

The equation for potential energy is:

E = w h

where h indicates the height that the person needs to ascend, therefore:

h = 3000 / 1067.52

h = 2.81 m

<span />

<span>
</span>

<span>Thus, he must ascend 2.81 meters</span>

3 0
2 months ago
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