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3 years ago

5

A fugitive tries to hop on a freight train traveling at a constant speed of 5.2 m/s . Just as an empty box car passes him, the f

ugitive starts from rest and accelerates at a = 1.3 m/s2 to his maximum speed of 6.1 m/s , which he then maintains. How long does it take him to catch up to the empty box car? What is the distance traveled to reach the box car? Express your answer to two significant figures and include the appropriate units.

1 answer:

RSB [31]3 years ago

6 0

Answer:

16 seconds

97 m

Explanation:

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{6.1-0}{1.3}\\\Rightarrow t=4.692\ s$

Train has moved a distance of $5.2\times 4.692=24.3984\ m$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 1.3\times 4.692^2\\\Rightarrow s=14.3096\ m$

Fugitive needs to run $24.3984-14.3096=10.0888\ m$ to make up

Relative speed of fugitive $6.1-5.2=0.9\ m/s$

Time taken

$t=\dfrac{10.0888}{0.9}\\\Rightarrow t=11.209\ s$

Time taken to catch the box = 4.692+11.209 = 15.901 seconds = 16 seconds

Distance traveled by the fugitive = $6.1\times 15.901=96.9961\approx 97\ m$

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3 years ago

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

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Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

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Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

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Answer:

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