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3 years ago

Compare and contrast the velocity at which X-rays and radio waves travel

2 answers:

8 0

Radio: longitudinal, light: transverse. x rays are produced by voltage and they are more penetrating Gamma Xrays Ultrasound Visible light Infrared Radiowaves. that is the order. colours are different since they absorb part of the colour spectrum & emit the rest

posledela3 years ago

6 0

<span> radio: longitudinal, light: transverse. x rays are produced by voltage and they are more penetrating Gamma Xrays Ultrasound Visible light Infrared Radiowaves. that is the order. colours are different since they absorb part of the colour spectrum & emit the rest </span>

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Who was the first woman to compete and win a championship in the Olympics

natita [175]

Hélène de Pourtalès she was the first women

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3 years ago

HEY CAN ANYONE PLS ANSWER DIS!!!!!!!!!!!!

Salsk061 [2.6K]

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3 years ago

You are listening to the radio when one of your favorite songs comes on, so you turn up the volume. If you managed to increase t

andrew-mc [135]

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

$\beta_{dB} = 10log_{10} \frac{I}{I_0}$

Where,

I = Acoustic intensity in linear scale

$I_0$ = Hearing threshold

The value in decibels is 17dB, then

$17dB = 10log_{10} \frac{I}{I_0}$

Using properties of logarithms we have,

$\frac{17}{10} = log_{10} \frac{I}{I_0}$

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$\frac{I}{I_0} = 10^{1.7}$

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3 years ago

A 58 kg skier is going down a 35 degree slope. The areaof each

maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

$N = mgcos\theta$

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$N' = \frac{mgcos\theta}{2}$

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$P = \frac{F}{A}$

Replacing F with N'

$P = \frac{N'}{A}$

$P = \frac{\frac{mgcos\theta}{2}}{A}$

Our values are given as,

$m = 58kg$

$g = 9.8m/s^2$

$\theta = 35\°$

$A = 0.3m^2$

Replacing we have that

$P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}$

$P = 776.01Pa$

Therefore the pressure exerted by each ski on the snow is 776.01Pa

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3 years ago

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