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ale4655 [162]
3 years ago
7

The engineer of a passenger train traveling at 25.0m/s sights a freight train whose caboose is 200m ahead on the same track. The

freight train is traveling at 15.0 m/s in the same direction as the passenger train. The engineer of the passenger train immediatly applies the brakes causing a constant acceleration of .100m/s^2 in a direction opposite to the trains velocity, while the freight train continues with constant speed. will there be a collision?
Physics
1 answer:
Vika [28.1K]3 years ago
5 0

Answer:

No, there won't be a collision.

Explanation:

We will use the constant acceleration formulas to calculate,

v = u + a*t

0 = 25 + (-0.1)*t

t = 250 seconds (the time taken for the passenger train to stop)

v^2 = u^2 + 2*a*s

0 = (25)^2  + 2*(-0.1)*s

s = 3125 m (distance traveled by passenger train to stop)

If the distance traveled by freight train in 250 seconds is less than (3125-200=2925 m) than the collision will occur

Speed*time = distance

Distance = (15)*(250)

Distance = 3750 m

As the distance is way more, there won’t be a collision

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A car drives around a curve with radius 539 m at a speed of 32.0 m/s. The road is banked at 5.00°. The mass of the car is 1.40 ×
HACTEHA [7]

Answer:

f_r = 150.47 N

Explanation:

given,

r = 539 m

v = 32 m/s

road banked at = 5°

∑ F_x

\dfrac{mv^2}{r}= N sin \theta + f_r cos \theta

∑ F_y = 0

0 = N cos \theta - f_r sin \theta - mg

N = \dfrac{f_rsin \theta + mg}{cos \theta}

\dfrac{mv^2}{r}= (\dfrac{f_rsin \theta + mg}{cos \theta})sin \theta + f_r cos \theta

              = f_r sin \theta tan \theta + mg tan \theta + f_r cos \theta

        f_r = \dfrac{\dfrac{mv^2}{r}- mg tan\theta}{sin\theta tan \theta + cos \theta}

         f_r = \dfrac{\dfrac{1.4\times 10^3 \times 32^2}{539}- 1.4\times 10^{3}\times 9.8 \times 0.087}{0.087 \times 0.087 + 0.996}

f_r = 150.47 N

8 0
3 years ago
Two people are each
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Answer:

the rope should break

Explanation:

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3 0
2 years ago
Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo
sergiy2304 [10]

Answer:

a=9\ cm/s^2

Explanation:

<u>Average Acceleration </u>

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When t_2=4\ sec, the graph shows a value of v_2=36\ cm/s

When t_1=0\ sec, the object is at rest, v_1=0

We compute the average acceleration as

\displaystyle a=\frac{v_2-v_1}{t_2-t_1}

\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}

\displaystyle a=\frac{36\ cm/s}{4\ s}

\boxed{a=9\ cm/s^2}

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Alenkasestr [34]

Answer: Newton, the unit of force, is defined based on Newton's Second Law (F=ma), as the force required to give a mass of one kilogram an acceleration of 1 meter/second2. Thus, it is derived from these other units.

Explanation:

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