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2 years ago

Using the data provided below, calculate the corrected wavelength for a spectroscope reading of 6.32.

Physics

1 answer:

weqwewe [10]2 years ago

6 0

Answer:

646.6 nm

Explanation:

Using the data given, we fit a linear model :

Spectroscope Readings Known Helium Wavelengths (nm)

3.65 388.9

4.75 486.6

4.90 501.6

5.65 587.6

6.45 667.8

7.00 706.5

Using technology like excel to fit the model, the regression model obtained is :

Y = 97.96582X + 27.48458

Where, y is the predicted or corrected wavelength value.

This mod could be used to calculate the corrected wavelength for a given Spectroscope value :

Given a Spectroscope value of 6.32 ; the corrected wavelength is obtained by replacing x in the equation by 6.32 and calculate y ;

Y = 97.96582X + 27.48458

x = 6.32

Y = 97.96582(6.32) + 27.48458

Y = 619.1439824 + 27.48458

Y = 646.6285624

Corrected wavelength value is : 646.6

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A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,

Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

$\sum F_x = ma_x\\$

$Fm - Ff = ma_x\\$ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

$Ff = \mu R\\Ff = \mu mg\\$

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

$Fm - Ff = ma_x\\Fm - \mu mg = ma\\$

Substitute the given parameters

$Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a = 1.736m/s^2\\$

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0

3 years ago

Differences between looping and simersaulting

Anika [276]

Somersaulting- for longer distances.It bends the narrow end in the direction it wants to go & takes grip with tentacles. It releases the broad end and straightens up. like this it continues. looping- for shorter distances.

Hope this helps

3 0

3 years ago

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

DochEvi [55]

Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

d) $K.E=0.096J$

e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

Therefore

$V_m=A\omega\\\\V_m=8.1*0.1$

$V_m=0.81m/s$

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

$a_m=\omega^2A$

$a_m=8.1^2*0.1$

$a_m=6.561m/s^2$

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*0.3*0.8^2$

$K.E=0.096J$

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

$T=\frac{2*3.142}{8.1}$

$T=0.78sec$

Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

$F=1.29sec$

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

Where

'= signify order of differentiation

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vovikov84 [41]

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Mademuasel [1]

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