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3 years ago

Would the energy of the wave increase or decrease if the speed of the wave increases? Why?

1 answer:

4 0

Answer: The energy of a wave increases with an increase in the amplitude of the wave. The energy of a wave decreases with an increase in the amplitude of a wave.

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The vibrations along a transverse wave move in a direction _________.

GrogVix [38]

Answer: perpendicular to it oscillations.

Explanation: A transverse wave is a wave whose oscillations is perpendicular to the direction of the wave.

By perpendicular, we mean that the wave is oscillating on the vertical axis (y) of a Cartesian plane and the vibration is along the horizontal axis (x) of the plane.

Examples of transverse waves includes wave in a string, water wave and light.

Let us take a wave in a string for example, you tie one end of a string to a fixed point and the other end is free with you holding it.

If you move the rope vertically ( that's up and down) you will notice a kind of wave traveling away from you ( horizontally) to the fixed point.

Since the oscillations is perpendicular to the direction of wave, it is a transverse wave

5 0

3 years ago

****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)

notka56 [123]

but I think it's a and f

and

b and e

3 0

3 years ago

A 15-kg block at rest on a horizontal frictionless surface is attached to a very light ideal spring of force constant 450 N/m. T

den301095 [7]

Answer:

0.266 m

Explanation:

Assuming the lump of patty is 3 Kg then applying the principal of conservation of linear momentum,

P= mv where p is momentum, m is mass and v is the speed of an object. In this case

$m_pv_p=v_c(m_p+m_b)$ where sunscripts p and b represent putty and block respectively, c is common velocity.

Substituting the given values then

3*8=v(15+3)

V=24/18=1.33 m/s

The resultant kinetic energy is transferred to spring hence we apply the law of conservation of energy

$0.5(m_p+m_b)v_c^{2}=0.5kx^{2}$ where k is spring constant and x is the compression of spring. Substituting the given values then

$(3+15)*1.33^{2}=450*x^{2}\\x\approx 0.266 m$

7 0

3 years ago

List 5 possible effects of not adhering to standards of measurement

murzikaleks [220]

Answer:

factual evidence of customer-service levels.

better understanding of cross-functional performance.

enhanced alignment of operations with strategy.

evidence-based determination of process improvement priorities.

detection of performance trends.

better understanding of the capability range of a process.

3 0

3 years ago

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

pav-90 [236]

A) $5.0\cdot 10^{-11} m$

The energy of an x-ray photon used for single dental x-rays is

$E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J$

The energy of a photon is related to its wavelength by the equation

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelength

Re-arranging the equation for the wavelength, we find

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m$

B) $2.0\cdot 10^{-11} m$

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

$E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J$

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m$

4 0

3 years ago