At the same amplitude or volume, which sound would have the highest frequency?

vesna_86 [32]

80 meters should be it

3 0

4 years ago

You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago

1) Construct a graph showing the relationship between and independent variable and a dependent variable. 2) Demonstrate potentia

lana66690 [7]

Answer:

lolfsedddddddddddddd

Explanation:

8 0

4 years ago

Read 2 more answers

Which is true about atoms?

tiny-mole [99]

Answer:

A

Explanation:

6 0

3 years ago

Read 2 more answers

Can someone please help?? I don’t understand this material!!!

Debora [2.8K]

Answer:

1)

When the person throws the ball away, the person rolls backward. This is due to the law of conservation of momentum: in fact, the total momentum of the person+ball system must be conserved.

At the beginning,

$p_i=0$

after throwing the ball, the total momentum is the sum of the momentum of the person and of the ball:

$p_f=p_p + p_b$

Since momentum is conserved,

$p_i = p_f\\0=p_p+p_b$

So

$p_p = -p_b$

Therefore, the person has equal momentum (in magnitude) but opposite direction to the ball, so the person rolls backward.

However, if the person hold to the ball, then they will have same momentum (moving in the same direction). In order to conserve the total momentum (which was zero at the beginning), the only possible solution is that

$p_p=p_b=0$

which means that both the person and the ball will remain at rest. This is because there are no external forces acting on the system, so the system cannot move.

2)

The change in momentum of an object is given by

$\Delta p=m(v-u)$

where

m is the mass of the object

v is its final velocity

u is the initial velocity

For the clay ball in this problem, we have:

m = 50 g = 0.050 kg

v = 0 m/s (it sticks on the wall)

u = 1 m/s

So its change in momentum is

$\Delta p_c=(0.050)(0-1)=-0.050 kg m/s$

For the superball, we have:

m = 50 g = 0.050 kg

v = -0.8 m/s (it bounces back)

u = 1 m/s

So its change in momentum is

$\Delta p_s = (0.050)(-0.8-1)=-0.09 kg m/s$

So, the superball has a greater change in momentum (in magnitude).

3a)

According to Newton's third law of motion:

"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A".

Here, we have a Hummer and a Beetle colliding head-on: we can identify them as object A and object B. Therefore, according to Newton's third law:

- The action force is the force of impact exerted by the Hummer on the Beetle

- The reaction force is the force of impact exerted by the Beetle on the Hummer

And according to the Law, the two forces are equal in magnitude: so, the two vehicles experience the same force of impact.

3b)

The change in momentum of each vehicle during the collision can be written as

$\Delta p = F\Delta t$ (1)

where

$\Delta p$ is the change in momentum

$F$ is the force experienced by the vehicle

$\Delta t$ is the duration of the collision

in part 3a), we said that the two vehicles experience the same force in the collision.

Moreover, the duration of the collision, $\Delta t$ , is the same for the two vehicles.

As a result, according to formula (1), the two vehicles have same change in momentum (however, the directions would be opposite, since they experience force in opposite directions).

3c)

According to Newton's second law of motion, the acceleration of an object is given by:

$a=\frac{F}{m}$

where

F is the force experienced by the object

m is its mass

a is its acceleration

In part 3a), we stated that the force experienced by the Beetle and the Hummer is the same. However, the mass of the Beetle is smaller than the mass of the Hummer: from the equation we see that the acceleration is inversely proportional to the mass, therefore the Beetle will experience a greater acceleration.

4a)

The force experienced by the dashboard on the car is given by:

$F=\frac{\Delta p}{\Delta t}$

Where

$\Delta p$ is the change in momentum

$\Delta t$ is the duration of the collision

In a padded dashboard, the duration of the collision $\Delta t$ is larger than the duration of the collision for a hard dashboard. According to the equation above, the force experienced by the dashboard (and so, the car) is inversely proportional to the duration of the collision: therefore, since the padded dashboard has a larger $\Delta t$ , it will experience a smaller force than the hard dashboard.

4b)

The force experienced by the climber if falling is given by

$F=\frac{\Delta p}{\Delta t}$

Where

F is the force experienced by the climber

$\Delta p$ is his change in momentum

$\Delta t$ is the duration of fall

Nylon is a very elastic material, so it is able to "soften" the fall by stretching a lot. As a result, the nylon increases the value of $\Delta t$ in the formula. Since the force experienced by the climber is inversely proportional to $\Delta t$ , the climber will feel less force thanks to the nylon.

4c)

This technique is used to exploit the "push" given by the second car of the train to the first car when the brakes are applied.

At first, the engine is started, and the first car starts accelerating, pulling the second car (and the following cars). Then, the brakes are applied on the first car: however, the second car keeps moving by inertia, so then it gives a push forward on the first car. Then, this action is repeated several times, so that this push exerted by the second car is exploited several times.

3 0

3 years ago