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4 years ago

How do Air molecules enable sound to travel from a radios speaker to your ears?

2 answers:

6 0

The radio frequencies push one air molecule that then bumps into a different air molecule.....which then hits another and another causing a line of crashing molecules that lead inside your ear and hits your ear drum causing it to vibrate which causes the sounds.

madreJ [45]4 years ago

6 0

The molecules bounce and colide and hit like dominoes and vibrate and we haer it

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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

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

An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro

Mashutka [201]

1) The distance travelled by the electron is $1\cdot 10^{17} m$

2) The time taken is $4.0\cdot 10^{10}s$

Explanation:

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

$u=5\cdot 10^6 m/s$ is the initial velocity

v = 0 is the final velocity (it comes to a stop)

$a=-1.25\cdot 10^{-4} m/s^2$ is the acceleration

Solving for s, we find the distance travelled:

$s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m$

The total time taken for the electron in its motion can also be found by using another suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

$u=5\cdot 10^6 m/s$

v = 0

$a=-1.25\cdot 10^{-4} m/s^2$

And solving for t, we find the time taken:

$t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0

3 years ago

The Nardo ring is a circular test track for cars. It has a circumference of 12.5 km. Cars travel around the track at a constant

KIM [24]

Answer: 50 km, 0, 27.78 m/s

Explanation:

Given

Circumference of the track is $12.5\ km$

Speed of car is $100\ km/h$

Car drives for $30\ \text{minute}\ or\ 0.5\ hr$

(a)Distance traveled is

$\Rightarrow D=100\times 0.5\\\Rightarrow D=50\ km$

(b)displacement of the car

It can be observed that 12.5 is a multiple of 50, that is, 50 km can be interpreted as 4 complete rounds of the track.

Therefore, the displacement of the car is zero.

(c)To convert kmph to m/s, multiply the entity by $\frac{5}{18}$

$\Rightarrow 100\times \dfrac{5}{18}\\\\\Rightarrow 27.78\ m/s$

5 0

3 years ago

Steel Usually forms a

meriva

Answer:

Permanent magnetism (of the steel)

make me brainliestt :))

6 0

4 years ago

The point of origin of an epileptic seizure is called the ____.â

djyliett [7]

The point of origin of an epileptic seizure is called the focus. It is the part of the brain that is affected by epilepsy. This is a term used for people with partial epilepsy. The focus can be the frontal lobe, occipital lobe, mesial temporal lobe or the parietal lobe. During seizures, it is the that specific part of the brain that is being effected. Seizures happen when a surge in the electrical activities happen on certain parts. Symptoms are contractions of the muscles, blackouts and visual disturbances. Seizures can be cured and, for partial epilepsy, they are treated with AEDs or antiepileptic drugs.

3 0

3 years ago