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liubo4ka [24]
3 years ago
11

Engineers are investigating the properties of a material for use as a wrapping product. Three identical

Physics
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

1st

Explanation:

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Earthquakes <br> Shaking of the ground <br> Seismographs <br> Scientific method
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C. seismographs

Explanation:

Theirs another name for seismographs but c is correct

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What factors in the atmosphere seem to have the greatest effect in the amount of radiation received from the sun?
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The ozone layer seem to have the greatest affect on the amount of radiation received from the sun.
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A rock is dropped off a cliff and falls the first half of the distance to the ground in t1 seconds. If it falls the second half
Hitman42 [59]

Answer:

t2/t1 = \sqrt{2} - 1

Explanation:

The expression for the second law of motion is given below:

h = ut + 0.5at^2

<u>For first half distance</u>

Object is initially at rest, so its initial speed u = 0

Object falls at half the distance, so h = h/2 where t = t1

Hence, we have

h/2 = at1^2/2 - equation 1

<u>For second half distance: </u>

Similarly,

h = a(t1 + t2)^2/2 - equation 2 where t = t1 + t2 and u= 0

Using equation 2 by equation 1

we obtain 2 = (t1 + t2)^2/t1^2

Hence t2/t1 + 1 = \sqrt{2}

Hence t2/t1 = \sqrt{2} - 1

5 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
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