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ivolga24 [154]
2 years ago
3

A volume charge distribution in free space is characterized by the density rho v = q ____ 2Ad exp (− |z| / d ) where d is a dist

ance along z, A is the area of a surface parallel to the xy plane, and q is a fixed charge quantity. The charge distribution exists everywhere. (a) Find the electric field intensity, E, everywhere. (b) What is the interpretation of q?

Physics
1 answer:
N76 [4]2 years ago
6 0

Answer:

+ for z>0 and - for z<0

Explanation:

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Stu wanted to calculate the resistance of a light bulb connected to a 4.0 V battery, with a resulting current of 0.5 A. He used
Dimas [21]
His answer was incorrect because according to ohm's law the formula used should have been R=V/I instead of multiplying and the answer should be 8ohms
4 0
2 years ago
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You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.
Afina-wow [57]

Answer:

<em>a) 17.05 mph</em>

<em>b) 54.7°  northeast direction</em>

<em>c) 10.71 mph</em>

<em>The direction is -22.58° relative to the east.</em>

<em></em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

V_{y} = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

V_{x} = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{13.89^{2} +9.89^{2}   } = <em>17.05 mph  This is your airspeed</em>

b) To get your direction, we use

tan ∅ = V_{y} /V_{x}

tan ∅ = 13.89/9.89 = 1.413

∅ = tan^{-1}(1.413) = <em>54.7°  northeast direction</em>

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

V_{y} = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = \sqrt{V^{2} _{y}+V^{2} _{x}  }

==> \sqrt{4.11^{2} +9.89^{2}   } = <em>10.71 mph  This is your airspeed</em>

Your direction will be,

tan ∅ = V_{y} /V_{x}

tan ∅ = -4.11/9.89 = -0.416

∅ = tan^{-1}(-0.416) =<em> -22.58°  this is the angle you'll travel relative to the east.</em>

<em>To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.</em>

5 0
2 years ago
A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, t
LenKa [72]

Answer:

a) W = - 318.26 J, b)  W = 0 , c) W = 318.275 J , d) W = 318.275 J , e) W = 0

Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

    fr - Wₓ = 0

    fr = mg sin 60

    fr = 15 9.8 sin 60

    fr = 127.31 N

Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

       W = 318.275 J

e) the net force

    F ’= fr - Wx = 0

    W = F ’L cos 0

    W = 0

4 0
2 years ago
A 20 watt lightbulb uses 20 Joules of energy every second.A person expends 50 watts of energy per stair when climbing up stairs.
Pachacha [2.7K]
20W = 20 J/s

Energy expended during climbing stairs = 50 W of energy/stair = 50J/stair

For 20 stairs, Total energy = 50x20 = 1000 J

This can light bulbs for, T= 1000J/20 J/s =50 seconds
6 0
2 years ago
A kite is 100m above the ground. If there are 200m of string out, what is the angle between the string and the horizontal? (Assu
belka [17]

Answer:

the answer is 30°

Explanation:

due to:

sin law of sines

\frac{sin 90}{200} =\frac{sin\beta }{100}\\arcsin(100\frac{sin90}{200} )= 30°

3 0
2 years ago
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