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3 years ago

9

Qualitative studies draw conclusions. True False

1 answer:

Marina86 [1]3 years ago

3 0

Answer:

Hey there!

This is false. A qualitative study is about how a thing looks, not based on any mathematical or scientific data. Quantitative studies, on the other hand, draw conclusions.

Let me know if this helps :)

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Consider a Young's two-slit experiment in which the wavelength of light is 37% smaller than the distance between the slits. How

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Answer: The answer is 3

Explanation:

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Viefleur [7K]

(3) The frictional force exerted by the floor on the box

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Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

vredina [299]

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44$\: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$

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3 years ago

An instrument is defective for all of the following conditions EXCEPT:_______

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Answer:

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Consider the nearly circular orbit of Earth around the Sun as seen by a distant observer standing in the plane of the orbit. Wha

ikadub [295]

We have that the spring constant is mathematically given as

$k=2.37*10^{11}N/m$

Generally, the equation for angular velocity is mathematically given by

$\omega=\sqrt{k}{m}$

Where

k=spring constant

And

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

Therefore

$\frac{2\pi}{T}=\sqrt{k}{n}$

Hence giving spring constant k

$k=m((\frac{2 \pi}{T})^2$

Generally

Mass of earth $m=5.97*10^{24}$

Period for on complete resolution of Earth around the Sun

$T=365 days$

$T=365*24*3600$

Therefore

$k=(5.97*10^{24})((\frac{2 \pi}{365*24*3600})^2$

$k=2.37*10^{11}N/m$

In conclusion

The effective spring constant of this simple harmonic motion is

$k=2.37*10^{11}N/m$

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