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lisabon 2012 [21]

3 years ago

10

9. If the musician hit the drum on a stage, how would the sound wave behave differently if he hit it the drum if the drum were s

ubmerged in a swimming pool? You may use pictures to help with your explanation. Be very thorough with your answer! If you use any sources other than your brain to help you with this question, be sure to list them!

1 answer:

egoroff_w [7]3 years ago

4 0

I think it would be yes because the drum is submerged in water and the water would slow the sound waves, making the sound softer. Right?

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Which factors affect gravitational potential force but not elastic potential energy

Inga [223]

Answer:

The mass of the object involved and the value of the gravitational acceleration

Explanation:

- Gravitational potential energy is defined as the energy possessed by an object in a gravitational field due to its position with respect to the ground:

$U=mgh$

where m is the mass of the object, g is the gravitational acceleration and h is the heigth of the object with respect to the ground.

- Elastic potential energy is defined as the energy possessed by an elastic object and it is given as:

$U=\frac{1}{2}kx^2$

where k is the spring constant of the elastic object, while x is the compression/stretching of the spring with respect to the equilibrium position.

As we can see from the equations, both types of energy depends on the relative position of the object/end of the spring with respect to a certain reference position (h in the first formula, x in the second formula), but gravitational potential energy also depends on m (the mass) and g (the gravitational acceleration) while the elastic energy does not.

3 0

3 years ago

Read 2 more answers

Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

$g=9.8 m/s^2$ is the acceleration of gravity

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

C)

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0

3 years ago

For a moving object, the force acting on the object varies directly with the object's acceleration. When a force of 81 N acts on

tensa zangetsu [6.8K]

Answer:

18 N

Explanation:

You have to do:

81 N: 9 ms2= x: 2 ms2

6 0

3 years ago

A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force of friction on the cart

Ber [7]

Newton's second law states that the resultant of the forces applied to an object is equal to the product between the object's mass and its acceleration:
$\sum F = ma$
where in our problem, m is the mass the (child+cart) and a is the acceleration of the system.

We are only concerned about what it happens on the horizontal axis, so there are two forces acting on the cart+child system: the force F of the man pushing it, and the frictional force $F_f$ acting in the opposite direction. So Newton's second law can be rewritten as
$F-F_a = ma$
or
$F=ma + F_f$

since the frictional force is 15 N and we want to achieve an acceleration of $a=1.50 m/s^2$ , we can substitute these values to find what is the force the man needs:
$F=(30 kg)(1.5 m/s^2)+15 N=60 N$

8 0

3 years ago

A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

$A=\pi r^2$

$A =\pi (0.2)^2$

$A =0.1256 m^2$

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

$\mu=IA$

$\mu=2(0.1256)$

$\mu=0.25 A\cdot m^2$

Therefore the magnetic dipole moment is $0.25A\cdot m^2$

6 0

3 years ago