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4 years ago

Which option was changed in the "Waves" lab to see how it would affect wave speed?

Physics

2 answers:

Ludmilka [50]4 years ago

7 0

Answer:

Option-(B): How hard the string was plucked

Explanation:

<u>Observing the speed of light in a string:</u>

The speed,s of the waves generated from a source can be analyzed well by making some number of changes but to know how hard the string was plucked, we will get an idea of the medium's flexibility or its ability to move in a more smooth wave as the energy,E transfer occurs through it. The energy,E transfer through the string is mainly considered as more of a longitudinal waves.

We know well about the transfer of energy in the longitudinal forms, as it has two main regions the rarefaction and the compression. When the energy is more or in dense form it is the point of compression, while to make some changes in the position of the string plucked in can be very easily analyzed by the observer. As the position or point where the string is plucked can be a very distinguishing factor due to the fact that angle between the destination and the receiving end matters a lot in the wave propagation.

gizmo_the_mogwai [7]4 years ago

4 0

Answer:

B. how hard the string was plucked

Explanation:

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answer is 36

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3 years ago

Decreasing the trains velocity will ______ the kinetic energy.<br><br> Decrease or Increase

alex41 [277]

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3 years ago

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e

Sati [7]

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

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ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

5 0

4 years ago

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m

Anna35 [415]

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

$I = \frac{P}{4\pi r^2},$

Where

P = Power

r = Radius

Replacing the values that we have,

$I = \frac{60}{(4*\pi (0.7)^2)}$

$I = 9.75 Watt/m^2$

The definition of intensity tells us that,

$I = \frac{1}{2}\frac{B_o^2 c}{\mu}$

Where,

$B_0 =$ Magnetic field

$\mu =$ Permeability constant

c = Speed velocity

Then replacing with our values we have,

$9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}$

Re-arrange to find the magnetic Field B_0

$B_o = 2.86*10^{-7} T$

Therefore the amplitude of the magnetic field of this light is $B_o = 2.86*10^{-7} T$

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3 years ago

A thin glass rod is submerged in oil. What is the critical angle for light traveling inside the rod? The index of refraction for

Alekssandra [29.7K]

When light travels from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle of incidence above which light is reflected only (no refraction occurs), and the value of this critical angle is given by
$\theta_c = \arcsin ( \frac{n_2}{n_1} )$
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.

In this problem, the first medium is the glass ( $n_1 = 1.50$ ), while the second medium is oil ( $n_2 =1.46$ ), therefore the critical angle is given by
$\theta_c = \arcsin( \frac{1.46}{1.50} )=\arcsin(0.973)=76.7^{\circ}$

7 0

4 years ago