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3 years ago

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If your brother changes the channel, you will change it back, so then he will change it, then you will change it, etc, etc. This

situation is a metaphor for which of Newton's laws of motion?
-The law of inertia
-The law of action-reaction
-The law of acceleration
-None of the above

1 answer:

KonstantinChe [14]3 years ago

7 0

The answer is the law of inetia

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A violin string is 45.0 cm long and has a mass of 0.242 g. When tightened on the neck of the violin, the distance between the pi

stiks02 [169]

Answer:

The tension is 75.22 Newtons

Explanation:

The velocity of a wave on a rope is:

$v=\sqrt{\frac{TL}{M}}$ (1)

With T the tension, L the length of the string and M its mass.

Another more general expression for the velocity of a wave is the product of the wavelength (λ) and the frequency (f) of the wave:

$v= \lambda f$ (2)

We can equate expression (1) and (2):

$\sqrt{\frac{TL}{M}}$ = $\lambda f$

Solving for T

$T= \frac{M(\lambda f)^2}{L}$ (3)

For this expression we already know M, f, and L. And indirectly we already know λ too. On a string fixed at its extremes we have standing waves ant the equation of the wavelength in function the number of the harmonic $N_{harmonic}$ is:

$\lambda_{harmonic}=\frac{2l}{N_{harmonic}}$

It's is important to note that in our case L the length of the string is different from l the distance between the pin and fret to produce a Concert A, so for the first harmonic:

$\lambda_{1}=\frac{2(0.425m)}{1}=0.85 m$

We can now find T on (3) using all the values we have:

$T= \frac{2.42\times10^{-3}(0.85* 440)^2}{0.45}$

$T=75.22 N$

3 0

3 years ago

A vehicle travels from a 30m marker to a 100m marker. What is the change in distance?

sesenic [268]

The change in distance is 30 because if you subtract both number you'll get 30

3 0

3 years ago

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left

Dahasolnce [82]

Answer:

$x_c= \dfrac{5}{9}L$

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any distance x from point A mass density

$\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x$

$\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x$

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

$dI=\lambda x^2dx$

So total moment of inertia

$I=\int_{0}^{L}\lambda x^2dx$

By putting the values

$I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx$

By integrating above we can find that

$I=\dfrac {7}{12}\lambda_ 0 L^3$

Now to find location of center mass

$x_c = \dfrac{\int xdm}{dm}$

$x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}$

Now by integrating the above

$x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}$

$x_c= \dfrac{5}{9}L$

So mass moment of inertia $I=\dfrac {7}{12}\lambda_ 0 L^3$ and location of center of mass $x_c= \dfrac{5}{9}L$

8 0

3 years ago

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn

kumpel [21]

Answer:

a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is:

$\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}$

$\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]$

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

$\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}$

$\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}$

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

$\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }$

$\tan \alpha = 1.902$

$\alpha = \tan^{-1}1.902$

$\alpha \approx 62.266^{\circ}$

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0

2 years ago

What is the average speed of a kangaroo that hops 60 m in 5 s ?

jok3333 [9.3K]

The formula for average speed is
$v_{avg}= \frac{total distance}{total time}$
So we can just substitute our data.
$v_{avg}= \frac{60}{5}=12 \frac{m}{s}$ - its the result

6 0

2 years ago

Read 2 more answers