When cell A sends a message to

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

xz_007 [3.2K]

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.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

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

(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

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

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

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

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

4 0

4 years ago

Earth turns on its axis about once every 24 hours. The Earth's equatorial radius is 6.38 x 106 m. If some catastrophe caused Ear

san4es73 [151]

To solve this problem it is necessary to apply the concepts related to the Period of a body and the relationship between angular velocity and linear velocity.

The angular velocity as a function of the period is described as

$\omega = \frac{2\pi}{T}$

Where,

$\omega =$ Angular velocity

T = Period

At the same time the relationship between Angular velocity and linear velocity is described by the equation.

$v = \omega r$

Where,

r = Radius

Our values are given as,

$T = 24 hours$

$T = 24hours (\frac{3600s}{1 hour})$

$T = 86400s$

We also know that the radius of the earth (r) is approximately

$6.38*10^6m$

Usando la ecuación de la velocidad angular entonces tenemos que

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{86400}$

$\omega = 7.272*10^{-5}rad/s$

Then the linear velocity would be,

$v = \omega *r$

x $v = \omega *r$

$v= 463.96m/s$

The speed would Earth's inhabitants who live at the equator go flying off Earth's surface is 463.96

3 0

3 years ago

Why does the earth bulge at the equator?

sattari [20]

centrifugal force is a fictitious force. What is happening is that since the earth itself is not a rigid body it will deform when under motion. Although gravity attempts to make the earth spherical, as it is rotating the earth deforms, in such away that it flattens to become an oblique spheroid. This happens as the material at the equator must have a net resultant centripetal force (not centrifugal) which causes its position of equilibrium from the center of the earth to be further away than at the poles as they do not have this force as they are not rotating around the center of mass.

4 0

3 years ago

Uuse Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire loop. You will need to u

Alex777 [14]

Answer:

Explanation:

According to the Fleming's right hand rule, if we spread our right hand such that the thumb, fore finger and the middle finger are mutually perpendicular to each other, then the thumb indicates the direction of force, fore finger indicates the direction of magnetic field, then the middle finger indicates the direction of induced current.

According to the Lenz's law, the direction of induced emf is such that it always opposes the cause due to which it is produced.

4 0

3 years ago

Why is velocity most important in space launch?

Debora [2.8K]

Escape velocity is the velocity an object needs to escape the gravitational influence of a body if it is in free fall, i.e. no force other than gravity acts on it. Your rocket is not in free fall since it is using its thruster to maintain a constant velocity so the notion of "escape velocity" does not apply to it.

6 0

4 years ago

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