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vovikov84 [41]
3 years ago
5

Someone please help me

Physics
1 answer:
OLga [1]3 years ago
6 0
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A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
3 years ago
The Solar System is made up of eight planets, numerous comets, asteroids, and moons, and the Sun. The force that holds all of th
Nastasia [14]

Answer:

B. GRAVITY is the answer

7 0
3 years ago
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A golfer is on the edge of a 12.5 m bluff overlooking the 18th hole which is located 60 m from the base of the bluff. She launch
Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
Whats the force of a pitching machine on a baseball
ira [324]

Answer:

kinetic

Explanation:

8 0
3 years ago
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A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block c
schepotkina [342]

   After impact velocity = 14.968 ft/s

Weight and mass of Bullet and wooden block:

Bullet: w = 1oz = 1/16 lb m = 0.001941 lb

wooden block : W = 5lb M = 0.15528 lb

velocity of block and bullet immediately after impact:

Σmv1 + ΣImp = mv2

Resolving vertical component

( m× v₀cos30⁰) + 0 = ( m+M) v'

v' = ( m× v₀cos30⁰)/ (m+M)

v' = 14.968 ft/s

Horizontal and vertical component of the impulse exerted by block on the bullet:

   Here we will apply the principle of impulse and momentum.

 Horizontal component:

          -mv₀ cos30⁰ + RxΔt  =0

                                 RxΔt = mv₀sin30⁰

                                          = 0.001941 × 1400sin30⁰

                                   RxΔt = 1.3587 lb.s

         Vertical component:

                           -mv₀cos30⁰  + RyΔt =  -mv'

                                     RyΔt = m( v₀cos30⁰-v')

                                     RyΔt = 0.001941(1400cos30⁰ - 14.968)

                                              = 2.32 lb.s

     Learn more about impact here:

            brainly.com/question/15008937

                  #SPJ4  

8 0
2 years ago
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