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Nady [450]
3 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?
Physics
1 answer:
soldi70 [24.7K]3 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 ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
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Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

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v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

where v is the speed, T is the tension and \mu is the mass per unit length.

Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

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