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2 years ago

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A ski gondola is connected to the top of a hill by a steel cable of length 600 m and diameter 1.2 cm . As the gondola comes to t

he end of its run, it bumps into the terminal and sends a wave pulse along the cable. It is observed that it took 18 s for the pulse to return. Part A What is the speed of the pulse?

1 answer:

Gemiola [76]2 years ago

6 0

Answer:

v = 66.7 m/s

Explanation:

Given that,

The length of steel cable, L = 600 m

Diameter = 1.2 cm

It is observed that it took 18 s for the pulse to return.

The time taken to cover 600 m will be :

t = T/2

t = 9 s

Let v be the of the pulse. We know that,

$v=\dfrac{L}{t}\\\\v=\dfrac{600}{9}\\\\v=66.7\ m/s$

So, the speed of the pulse is equal to 66.7 m/s.

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Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

$Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\$

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

$F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\$

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2 years ago

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Answer:

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2 years ago

What is the main difference between the following two velocities: 7 m/s and -7m/s?

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2 years ago

The current theory of the structure of the

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Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

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$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

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valentinak56 [21]

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3 years ago