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lisabon 2012 [21]

3 years ago

10

What do wind and moving water have in common? Select one: They both have electrical energy They both have elastic potential ener

gy They both have electromagnetic energy (light) They both have kinetic energy

1 answer:

ozzi3 years ago

7 0

They both have kinetic energy because they are both moving and have force.
Hope this helped! :)

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Suppose a diving board with no one on it bounces up and down in a SHM with a frequency of 4.00 Hz. The board has an effective ma

kirza4 [7]

Answer:

I took 3*sqrt(10/83)= 1.110349815

And rounded to 1.11 Hz

Explanation:

7 0

2 years ago

A container contains 200g of water at initial temperature of 30°C. An iron nail of mass 200g at temperature of 50°C is immersed

d1i1m1o1n [39]

Answer:

Assuming there is no heat loss to the surrounding.

Heat lost by iron equals heat gained by water.

0.2(450)(50-x)=0.2(4200)(x-30)

x=31.94 °C

Explanation:

4 0

3 years ago

Read 2 more answers

A box has a momentum of

Kipish [7]

Answer:

67.9 kg*m/s

Explanation:

Pi = 38 kgm/s

F = 88.3N and ∆t = 0.338s

Final momentum Pf = Pi + F∆t = 38 + (88.3)(0.338) = 38 + 29.8454

=) Pf = 67.8454 kgm/s = 67.85kg*m/s

Your answer is 67.9kg*m/s with three significant figures

hope this helps your troubles!

6 0

2 years ago

ahrayia [7]

1. F = 8 Hz
2. V = 8 m/s
3. D = 8 m
4. T = 8 sec

5 0

2 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

2 years ago