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faust18 [17]
3 years ago
5

A 5.09 × 1014-hertz electromagnetic wave is traveling through a transparent medium. The main factor that determines the speed of

this wave is the
Physics
1 answer:
sergiy2304 [10]3 years ago
6 0
We are given an electromagnetic wave with a frequency of 5.09 x 10^14 Hz and travelling through a transparent medium. If the medium was vacuum, the speed of the wave would be equal to the speed of light. Otherwise, the main factor that would determine the speed of the wave is its wavelength.
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Zach, whose mass is 90 kg , is in an elevator descending at 12 m/s . the elevator takes 3.0 s to brake to a stop at the first fl
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Finding out the acceleration 12/3 = 4m/s^2
thus it is descending so the actual acceleration would be 9.8-4 = 5.8 m/s^2
the weight will be 90*5.8 = 522 N
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A person spins a tennis ball on a string in a horizontal circle (so that the axis of rotation is vertical). At the point indicat
Ilya [14]

Answer:

direction does the axis of rotation tilt toward after the blow is the z- direction

Explanation:

This is because

Due to the blow, there is a impulse imparted on the ball which gives a change in linear momentum () in the x direction

Thus When cross product of momentum in x direction is taking with r vector, we will get resultant in z direction. Hence change in angular momentum will be z direction.

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
What kind of tectonic plate boundary must have existed in the past where copper deposits are found, based on the way they are fo
Maksim231197 [3]

Answer:

older plates that are subducting almost perpendicularly, and when the location is far away from the edge of other tectonic plates, are the most likely areas for copper deposits to form

Explanation:

5 0
3 years ago
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