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2 years ago

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A tuning fork is sounded above a resonating tube (one end closed), which resonates at a length of 0.20 m and again at 0.60 m. If

the tube length were extended further, at what point will the tuning fork again create a resonance condition?

1 answer:

Nina [5.8K]2 years ago

6 0

To solve this problem we will apply the concepts related to resonance. The velocity with which sound travels in any medium may be determined if the frequency and the wavelength are known. Since the pipe is closed at one end, that produces frequencies ratio 1:3:5, then

$n_1 = \frac{V}{4L_1}$

Third resonance occurs at $5n_2$

$n_2 = \frac{V}{4L_2}$

At resonance $5n_2=n_1$

$5\frac{V}{4L_2} = \frac{V}{4L_1}$

$L_2 = 5L_1$

$L_2 = 5*0.2$

$L_2 = 1m$

Therefore at 1m will the tuning fork again create a resonance condition

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Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

sveta [45]

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with $m_{Pa}$ and its initial velocity with $u_{Pa}$

Let represent the mass of Ricardo with $m_{Ri}$ and its initial velocity with $u_{Ri}$

At rest ;

their velocities will be zero, i.e

$u_{Pa}$ = $u_{Ri}$ = 0

The initial momentum for this process can be represented as :

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = 0

after push off from each other then their final velocity will be $v_{Pa}$ and $v_{Ri}$

The we can say their final momentum is:

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

$m_{Pa}$ $u_{Pa}$ + $m_{Ri}$ $u_{Ri}$ = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

Since the initial velocities are stating at rest then ; u = 0

$m_{Pa}$ (0) + $m_{Pa}$ (0) = $m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$

$m_{Pa}$ $v_{Pa}$ + $m_{Ri}$ $v_{Ri}$ = 0

$m_{Pa}$ $v_{Pa}$ = - $m_{Ri}$ $v_{Ri}$

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So $m_{Ri} > m_{Pa}$ ;

Then $\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}$

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

$m_{Pa}$ $v_{Pa}$ = $m_{Ri}$ $v_{Ri}$

The ratio is

$\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1$

$v_{Pa} >v_{Ri}$

Therefore, Paula will have a greater speed than Ricardo after the push-off.

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A flat rectangular loop of wire carrying a 4.0-a current is placed in a uniform 0.60-t magnetic field. the magnitude of the torq

ZanzabumX [31]

Torque on a closed current carrying loop is given by formula

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here

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$\theta$ = angle between normal of the plane and magnetic field = 90 - 30 = 60 degree

Now we will have

$1.1 = 1* 4 * A * 0.60* sin60$

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So area of the loop is 0.53 m^2

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Will two separate 50db sounds together constitute a 100db sound explain mathematical

vfiekz [6]

Putting together two distinct 50 dB sound, do not create a 100 dB sound. Since decibels are logarithm of energy, creating two sounds together only makes the energy increase but the logarithm only goes up by somehow little. So increasing the sound by 10 dB, only makes it 10000 times louder because each 10 dB increase in sound makes the sound 10 times louder.

Twice as loud is an increase of 10Log (2) = 3.01 dB. So, 53,01 dB is twice as loud as 50dB.

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F1 = G*m²/D²
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2 years ago

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A slingshot can project a pebble at a speed as high as 38.0 m/s. (a) If air resistance can be ignored, how high (in m) would a p

kipiarov [429]

Answer:

73.67 m

Explanation:

If projected straight up, we can work in 1 dimension, and we can use the following kinematic equations:

$y(t) = y_0 + V_0 * t + \frac{1}{2} a t^2$

$V(t) = V_0 + a * t$ ,

Where $y_0$ its our initial height, $V_0$ our initial speed, a the acceleration and t the time that has passed.

For our problem, the initial height its 0 meters, our initial speed its 38.0 m/s, the acceleration its the gravitational one ( g = 9.8 m/s^2), and the time its uknown.

We can plug this values in our equations, to obtain:

$y(t) = 38 \frac{m}{s} * t - \frac{1}{2} g t^2$

$V(t) = 38 \frac{m}{s} - g * t$

note that the acceleration point downwards, hence the minus sign.

Now, in the highest point, velocity must be zero, so, we can grab our second equation, and write:

$0 m = 38 \frac{m}{s} - g * t$

and obtain:

$t = 38 \frac{m}{s} / g$

$t = 38 \frac{m}{s} / 9.8 \frac{m}{s^2}$

$t = 3.9 s$

Plugin this time on our first equation we find:

$y = 38 \frac{m}{s} * 3.9 s - \frac{1}{2} 9.8 \frac{m}{s^2} (3.9 s)^2$

$y=73.67 m$

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2 years ago