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r-ruslan [8.4K]

2 years ago

14

What happens when sound waves move faster?

1 answer:

Lady bird [3.3K]2 years ago

6 0

Hnjjjjj hejehtjriwje a wkirrhnemekwhr. Rthejb wine rnrie

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Energy is measured by joule(j)

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Can someone help me wit hthis science question?

klemol [59]

Answer:

see explanation

Explanation:

There is an increasing demand for materials and natural resources from a growing global population, especially those in more economically developed countries. The world's resources are being used up more quickly. The consumption of resources is spread unequally between MEDCs (more economically developed countries), who use more resources, and LEDCs (less economically developed countries), who use less.

The gap between the rich and poor is more evident when the resources are shared so unevenly and unfairly and natural resources like materials and natural energy cannot reach the demand of the people which can have consequences and be very difficult to manage. Having a lack of these materials in a country can result in prices going up for them, and the industry could be harder to work in because of a lack of materials.

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If the energy carried by a wave increases, which other wave property also increases?

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3 years ago

How does increasing the tension of a spring affect a wave on the spring?

Liono4ka [1.6K]

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You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.

Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

$F_{a} =\frac{F_{f}+F_{i} }{2}$ Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx : displacement

$W = F_{a} (x_{f} -x_{i} )$ :Formula (3)

$x_{f}$ : final deformation

$x_{i}$ :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

$F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N$

$F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N$

We calculate average force applying formula (2):

$F_{a} =\frac{15.4N+6.2N}{2} = 11 N$

We calculate the work done on the spring applying formula (3) : :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

$E_{pf} =\frac{1}{2} *k*x_{f}^{2}$

$E_{pi} =\frac{1}{2} *k*x_{i}^{2}$

$E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J$

$E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J$

W=ΔEp= 5.39 J-0.99 J = 4.4J

:

4 0

2 years ago