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Katen [24]
3 years ago
14

Ilya and Anya each can run at a speed of 7.50 mph and walk at a speed of 4.00 mph. They set off together on a route of length 5.

00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
Required:
a. How long does it take Anya to cover the distance of 5.00 ?
b. Find Anya's average speed.
c. How long does it take Ilya to cover the distance?
Physics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

T_A=0.93hours

S_A=5.23mph

T_l=0.87hour

Explanation:

From the question we are told that

Run speed of Ilya and Anya S_r=7.50mph

Walk speed of Ilya and Anya S_w=4.00mph

Total distance T_d=5.00miles

Anya Walks 1/2d and runs 1/2d

ilya walks 1/2t and runs 1/2d

Generally the distance walked by Ilay k is mathematically given by

Since ilya walks 1/2t and runs 1/2t

Therefore

\frac{k}{4}=\frac{5-k}{7.5}

k=1.7miles

Therefore llay runing distance is m

m=5-1.7\\m=3.3mile

a)Generally the Time walked by Anya T_A is mathematically given by

Anya Walks 1/2d and runs 1/2d

Therefore

t=\frac{2.5}{4.}

t_1=0.6hours

t=\frac{2.5}{7.5}

t_2=0.33hours

T_A=0.6+0.33\\

T_A=0.93hours

b)Generally the average speed of Anya S_A is mathematically given as

S_A=\frac{Distance}{T_t}

S_A=\frac{5.0}{0.93}

S_A=5.23mph

Generally the time taken b y llay T_l is mathematically given by

T_l=2t

T_l=2*(\frac{1.7}{4})

T_l=0.87hour

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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

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Remember KE=M*V2/2…
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Answer:

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Explanation:

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The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

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work done is equal to Force into displacement.

W = F . s            

W = 10 x 5              

W = 50 J                

Work done by the object when 10 N force is applied is equal to 50 J

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