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Fiesta28 [93]
3 years ago
5

A cattle train left Miami and traveled toward New York. 14 hours later a diesel train left traveling at 45 km/h in an effort to

catch up to the cattle train. After traveling for four hours the diesel train finally caught up. What was the cattle train's average speed?
Physics
1 answer:
PSYCHO15rus [73]3 years ago
7 0

Answer:

The cattle train´s average speed was 10 km/h.

Explanation:

Hi there!

The traveled distance of the trains can be calculated using this equation:

d = v · t

Where:

d = traveled distance.

v = velocity of the train.

t = time.

The diesel train travels 4 hours at 45 km/h, then, the traveled distance will be:

d = 45 km/h · 4 h = 180 km

Because the diesel train catches the cattle train after traveling 180 km, this will also be the distance traveled by the cattle train. But it took the cattle train (14 h + 4 h) 18 hours to travel that distance. Then, using the equation of distance:

d = v · t

180 km = v · 18 h

180 km/ 18 h = v

10 km /h = v

The cattle train´s average speed was 10 km/h

Have a nice day!

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Suppose we measure the energy stored in some inductor to be E when there is a current I running through it. If I double the curr
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If I double the current in the inductor, the new total energy will become 4E (option f).

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The coil or inductor is a passive component made of an insulated wire that stores energy in the form of a magnetic field due to its form of coiled turns of wire, through a phenomenon called self-induction. In other words, inductors store energy in the form of a magnetic field. The energy stored in the space where there is a magnetic field in the inductor is:

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where E is Energy [J], L is Inductance [H] and I is Current [A].

If you double the current in the inductor, then the new value of the current is I'= 2*I. So replacing the new total energy is:

E'=\frac{1}{2} *L*I'^{2}=\frac{1}{2} *L*(2*I)^{2}=\frac{1}{2} *L*4*I^{2}=4*\frac{1}{2} *L*I^{2}

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E'=4*E

<em><u>If I double the current in the inductor, the new total energy will become 4E (option f).</u></em>

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The minimum coefficient of static friction  between the pavement and the tires is 0.69.

The given parameters;

  • <em>radius of the curve, r = 90 m</em>
  • <em>angle of inclination, θ = 10.8⁰</em>
  • <em>speed of the car, v = 75 km/h = 20.83 m/s</em>
  • <em>mass of the car, m = 1100 kg</em>

The normal force on the car is calculated as follows;

F_n = mgcos(\theta)

The frictional force between the car and the road is calculated as;

F_k = \mu_k F_n\\\\F_k = \mu_k mgcos(\theta)

The net force on the car is calculated as follows;

mgsin(\theta) +  \mu_s mgcos(\theta) = \frac{mv^2}{r} \\\\mg(sin\theta \ + \ \mu_s cos\theta)= \frac{mv^2}{r} \\\\g(sin\theta \ + \ \mu_s cos\theta)= \frac{v^2}{r}\\\\sin\theta \ + \ \mu_s cos\theta = \frac{v^2}{rg}\\\\\mu_s cos\theta = sin\theta \  + \ \frac{v^2}{rg}\\\\\mu_s = \frac{sin\theta}{cos \theta} + \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(\theta) +   \frac{v^2}{cos (\theta)rg}\\\\\mu_s = tan(10.8) +  \frac{(20.83)^2}{cos(10.8) \times 90 \times 9.8} \\\\\mu_s = 0.19 + 0.5\\\\

\mu_s = 0.69

Thus, the minimum coefficient of static friction  between the pavement and the tires is 0.69.

Learn more here:brainly.com/question/15415163

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