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3 years ago

Which electromagnetic waves have the largest wavelength?

2 answers:

5 0

Answer:Red has the longest wavelength and violet has the shortest wavelength. When all the waves are seen together, they make white light. Ultraviolet (UV) light—is radiation with a wavelength shorter than that of visible light, but longer than X-rays, in the range 10 nm to 400 .

Explanation:

Gnoma [55]3 years ago

4 0

The electromagnetic waves with the longest wavelengths are the ones we call "radio waves".

They're anything with a wavelength of 1 millimeter or more. (Frequency of 300 GHz or less.)

That's longer than gamma rays, X-rays, ultraviolet, visible light, and infrared radiation.

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Newton's first law states that an object traveling with a constant velocity will remain traveling at a constant velocity unless

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C. The forces on an object traveling at terminal velocity are balanced.

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4 years ago

How is thermal energy transferred during conduction?

Irina-Kira [14]

Answer:

Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more. These molecules then bump into nearby particles and transfer some of their energy to them. This then continues and passes the energy from the hot end down to the colder end of the substance.

Explanation:

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3 years ago

When a wire loop is connected to a battery, what is produced in the loop?

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Answer:

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Explanation:

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3 years ago

The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are

r-ruslan [8.4K]

Answer with explanation:

Part a)

$v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s$

Part b)

$a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}$

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3 years ago

The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

kakasveta [241]

Answer:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

$Vector (OA) = 6i + 0j + 0k \\Vector (OB) = 0i + 3j + 2k \\Vector (OC) = 0i - 2j + 3k$

Next step is to find unit vectors $u_{AB} ,u_{AC}, u_{AD}, u_{AE}$ as follows:

$u_{AB} = \frac{vector(AB)}{magnitude(AB)} \\= \frac{OB - OA}{magnitude({vector(OB - OA))} }\\=\frac{-6i +3j+2k}{\sqrt{6^2 + 3^2+2^2} } \\\\=-0.857 i +0.429j+0.286k\\\\u_{AC} = \frac{vector(AC)}{magnitude(AC)} \\= \frac{OC - OA}{magnitude({vector(OC - OA))} }\\=\frac{-6i -2j-3k}{\sqrt{6^2 + 2^2+3^2} } \\\\=-0.857 i -0.286j+0.429k\\\\u_{AD} = +1i\\u_{AC} = -1k$

Using the diagram we find the corresponding vectors Forces:

$F_{AB} = F_{AB} i + F_{AB}j +F_{AB}k\\F_{AC} = F_{AC} i + F_{AC}j +F_{AC}k\\F_{AD} = F_{AD} i + F_{AD}j +F_{AD}k\\W = -160 k$

Equation of Equilibrium:

$Sum of forces = 0\\F_{AB}. u_{AB} + F_{AC}.u_{AC} + F_{AD}.u_{AD} + W = 0\\(-0.857F_{AB}i + 0.429F_{AB}j +0.286F_{AB}k) + (-0.857F_{AC}i - 0.286F_{AC}j +0.429F_{AC}k) + (+1F_{AD} i) + (-160k) = 0$

Comparing i , j and k components as follows:

$-0.857F_{AB} -0.857F_{AC} +1F_{AD} = 0\\+ 0.429F_{AB} - 0.286F_{AC} = 0\\+0.286F_{AB} +0.429F_{AC} = 160$

Solving Above equation simultaneously we get:

$F_{AB} = 172.1356\\F_{AC} = 258.2033\\F_{AD} = 368.8004$

3 0

4 years ago