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4 years ago

10

If the total mass hanging on one end of the string that creates the tension in the string is 0.300 kg , what is the mass density

(mass per unit length) μ of the string?

1 answer:

soldi70 [24.7K]4 years ago

7 0

Answer:

The mass density of the string is (0.3/L)kg/m

Explanation:

Mass density of the string = Mass/Length

Mass = 0.3kg

The length of the string is unknown so it is assumed to be L meter(s)

Therefore, mass deny of the string = 0.3kg/Lm = (0.3/L)kg/m

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A heavy rope 6.00 m long and weighing 29.4 N is attached at one end to a ceiling and hangs vertically. A 0.500-kg mass is suspen

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Answer:

a) $v=3.1252\ m.s^{-1}$

b) $v=39.0672\ m.s^{-1}$

c) $v=8.2685\ m.s^{-1}$

d) No,

No.

Explanation:

Given:

length of rope, $l=6\ m$

weight of the rope, $w=29.4\ N$

mass suspended at the lower end of the rope, $M=0.5\ kg$

<u>Now the mass of the rope:</u>

$m=\frac{w}{g}$

$m=\frac{29.4}{9.8}$

$m=3.01\ kg$

<u>So the linear mass density of rope:</u>

$\mu=\frac{m}{l}$

$\mu=\frac{3.01}{6}$

$\mu=0.5017\ kg.m^{-1}$

We know that the speed of wave in a tensed rope is given as:

$v=\sqrt{\frac{F_T}{\mu} }$

where:

$F_T=$ tension force in the rope

a)

At the bottom of the hanging rope we have an extra mass suspended. So the tension at the bottom of the rope:

$F_T=M\times g$

$F_T=0.5\times 9.8$

$F_T=4.9\ N$

Therefore the speed of the wave at the bottom point of the rope:

$v=\sqrt{\frac{4.9}{0.5017} }$

$v=3.1252\ m.s^{-1}$

b)

Tension at a point in the middle of the rope:

$F_T=M\times g+\frac{w}{2}$

$F_T=0.5\times 9.8+\frac{29.4}{2}$

$F_T=19.6\ N$

Now wave speed at this point:

$v=\sqrt{\frac{19.6}{0.5017} }$

$v=39.0672\ m.s^{-1}$

c)

Tension at a point in the top of the rope:

$F_T=M\times g+w$

$F_T=0.5\times 9.8+29.4$

$F_T=34.3\ N$

Now wave speed at this point:

$v=\sqrt{\frac{34.3}{0.5017} }$

$v=8.2685\ m.s^{-1}$

d)

Tension at the middle of the rope is not the average tension of tension at the top and bottom of the rope because we have an extra mass attached at the bottom end of the rope.

Also the wave speed at the mid of the rope is not the average f the speeds at the top and the bottom of the ropes because it depends upon the tension of the rope at the concerned points.

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Explanation:

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Answer:

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Assuming the gazelle stops, cheetah will run at 25.5 m/s - 21.6 m/s = 3.9 m/s to catch its prey (gazelle).

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