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kompoz [17]
2 years ago
6

Lindsey started biking to the park traveling 15 mph, after some time the bike got a flat so Lindsey walked the rest of the way,

traveling 4 mph. If the total trip to the park took 5 hours and it was 53 miles away, how long did Lindsey travel at each speed
Physics
1 answer:
-BARSIC- [3]2 years ago
5 0

Answer:

Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.

Explanation:

Speed = distance/time

Let the distance that Lindsey biked through be x miles and the time it took her to bike through that distance be t hours

Then, the rest of the distance that she walked is (53 - x) miles

And the time she spent walking that distance = (5 - t) hours

Her biking speed = 15 mph = 15 miles/hour

Speed = distance/time

15 = x/t

x = 15 t (eqn 1)

Her walking speed = 4 mph = 4 miles/hour

4 = (53 - x)/(5 - t)

53 - x = 4 (5 - t)

53 - x = 20 - 4t (eqn 2)

Substitute for X in (eqn 2)

53 - 15t = 20 - 4t

15t - 4t = 53 - 20

11t = 33

t = 3 hours

x = 15t = 15 × 3 = 45 miles.

(53 - x) = 53 - 45 = 8 miles

(5 - t) = 5 - 3 = 2 hours

So, it becomes evident that Lindsey biked 45 miles for 3 hours at 15 mph and walked 8 miles for 2 hours at 4 mph.

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Explanation:

A,D,E

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3 years ago
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

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2 years ago
. Friction is a rubbing force that ___________ a spinning yo-yo.
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elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0
2 years ago
a body starts from rest with a uniform acceleration of 2m s-2 find the distance covered by the body in 2s
Evgen [1.6K]
Finding acceleration= final velocity-initial velocity/ time taken (or A= V-U/T)

Final speed= 2m
Initial speed= 0m
Time taken= 2 seconds

2-0/2 so it’ll be 1m/s

2-0=0
2/2=

8 0
2 years ago
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