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3 years ago

Describe how the ringing sound of a telephone travels from the phone to your ear

Physics

1 answer:

ehidna [41]3 years ago

6 0

The phone gives off sound waves, and the sound waves go into your ear to be interpreted by ur brain

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If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later

garik1379 [7]

X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

5 0

3 years ago

8a. What is the equivalent resistance of the following circuit?

ollegr [7]

Answer: Take your pick

Explanation:

if they are all in parallel 1 /(1/100 + 1/300 + 1/50) = 30 Ω

if 50 is in parallel with 2 in series 1 / (1/(100 + 300) + 1/50) = 44.444...Ω

if 100 is in parallel with 2 in series 1 / (1/(50 + 300) + 1/100) = 77.777...Ω

if 300 is in parallel with 2 in series 1 / (1/(100 + 50) + 1/300) = 100 Ω

If 50 is in series with 2 in parallel 50 + 1/(1/100 + 1/300) = 125 Ω

If 100 is in series with 2 in parallel 100 + 1/(1/50 + 1/300) = 142.857...Ω

If 300 is in series with 2 in parallel 300 + 1/(1/50 + 1/100) = 333.333...Ω

If they are all in series 100 + 300 + 50 = 450 Ω

4 0

3 years ago

A fairground ride consists of a large vertical drum that spins so

expeople1 [14]

Answer:

v = 3.84 m/s

Explanation:

In order for the riders to stay pinned against the inside of the drum the frictional force on them must be equal to the centripetal force:

$Centripetal\ Force = Frictional\ Force\\\\\frac{mv^2}{r} = \mu R = \mu W\\\\\frac{mv^2}{r} = \mu mg\\\\\frac{v^2}{r} = \mu g\\\\v = \sqrt{\mu gr}$

where,

v = minimum speed = ?

g = acceleration due to gravity = 9.81 m/s²

r = radius = 10 m

μ = coefficient of friction = 0.15

Therefore,

$v=\sqrt{(0.15)(9.81\ m/s^2)(10\ m)}$

6 0

3 years ago

A starship blasts past the earth at 2.0*10^8 m/s .Just after passing the earth, the starship fires a laser beam out its back of

Vilka [71]

Answer:

at the speed of light ( $c=3.0\cdot 10^8 m/s$ )

Explanation:

The second postulate of the theory of the special relativity from Einstein states that:

"The speed of light in free space has the same value c in all inertial frames of reference, where $c=3.0\cdot 10^8 m/s$ "

This means that it doesn't matter if the observer is moving or not relative to the source of ligth: he will always observe light moving at the same speed, c.

In this problem, we have a starship emitting a laser beam (which is an electromagnetic wave, so it travels at the speed of light). The startship is moving relative to the Earth with a speed of 2.0*10^8 m/s: however, this is irrelevant for the exercise, because according to the postulate we mentioned above, an observer on Earth will observe the laser beam approaching Earth with a speed of $c=3.0\cdot 10^8 m/s$ .

7 0

3 years ago

In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo

irinina [24]

Answer:

The torque about the origin is $2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}$

Explanation:

Torque $\overrightarrow{\tau}$ is the cross product between force $\overrightarrow{F}$ and vector position $\overrightarrow{r}$ respect a fixed point (in our case the origin):

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

$\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]$

$\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}$

4 0

3 years ago