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3 years ago

The part of a longitudinal wave where the particles are spread apart is called a______?

Physics

2 answers:

Mnenie [13.5K]3 years ago

8 0

Answer:

Rarefaction (I think so..)

lilavasa [31]3 years ago

3 0

Answer:

In a longitudinal wave, particles of the medium vibrate in a direction that is parallel to the direction that the wave travels. Places where particles of the medium crowd closer together are called compressions. Places where particles of the medium spread farther apart are called rarefactions.

Explanation:

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SIL. TUILE I JUMU<br>Write down your own examples for each of the different effects of forces.

Harrizon [31]

Answer:

magnetic force

Explanation:

when some one gets charged up by maybe static eletricity the person will sustain the energy and the energy will make a force field around the person which can and will attract paper and he will be attractive

5 0

3 years ago

Read 2 more answers

Describe in terms of kinetic and potential energy what happens if an apple falls from a tree and comes to rest on the ground( wr

Lunna [17]

Answer:

An apple hanging at a branch has potential energy due its position. It can be written as PE= mgh where m is the mass of the apple h is the distance between the apple and the ground and g is the acceleration due to gravity.

as the apple falls from the tree it loses its potential energy and gains kinetic energy due to the movement of the apple. Its kinetic energy will be given by KE= 1/2mv² where m is the mass of the apple and v is the speed with which the apple falls.

As the apple falls the height or the distance reduces and PE becomes reduces. But it gains Kinetic energy due to its speed.

But when the apple falls to the ground and comes to rest its kinetic energy is converted to potential energy.

thus the total energy remains the same. it changes from one form to the other but remains unaltered.

6 0

3 years ago

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

3 years ago

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total di

S_A_V [24]

Answer:

1keff=1k1+1k2

see further explanation

Explanation:for clarification

Show that the effective force constant of a series combination is given by 1keff=1k1+1k2. (Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination. Also, each spring must exert the same force. Do you see why?

From Hooke's law , we know that the force exerted on an elastic object is directly proportional to the extension provided that the elastic limit is not exceeded.

Now the spring is in series combination

F $\alpha$ e

F=ke

k=f/e.........*

where k is the force constant or the constant of proportionality

k=f/e

$f_{eff} =f_{1} +f_{2}$ ............................1

also for effective force constant

divide all through by extension

1) Total force is

Ft=F1+F2

Ft=k1e1+k2e2

F = k(e1+e2) 2)

Since force on the 2 springs is the same, so

k1e1=k2e2

e1=F/k1 and e2=F/k2,

and e1+e2=F/keq

Substituting e1 and e2, you get

1/keq=1/k1+1/k2

Hint: For a given force, the total distance stretched by the equivalent single spring is the sum of the distances stretched by the springs in combination.

4 0

3 years ago

A women who normally weighs 400 N stands on top of a ladder so high she is one full additional earth radius above the surface of

AveGali [126]

The woman weighs 100N now.

Explanation:

Given:

F is given as 400N

one full additional earth radius formed

To Find:

Weight of the woman=?

Solution: