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neonofarm [45]
3 years ago
8

Which kinds of objects emit visible light in the electromagnetic spectrum?1)all objects2)radioactive objects3)relatively cold ob

jects4)relatively hot objects
Physics
1 answer:
Lostsunrise [7]3 years ago
8 0
Relatively hot objects emit visible light.
Some examples: 
==> the wire coils in the toaster; 
==> the spoon that you stuck in the flame on the stove;
==> the fine wire in the lightbulb when current goes through it.

VERY radioactive objects also do that.  But if you're actually
standing there watching an object that's THAT radioactive,
then you're in big trouble.
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To model time-variant data, one must create a new entity in an m:n relationship with the original entity.
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To model time-variant data, one must create a new entity in an m:n relationship with the original entity, is a False statement.

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11 months ago
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi
Bad White [126]
A) Agreed. 
<span>b) Value agreed but units should be W (watts). </span>

<span>c) Here's one method... </span>

<span>15 miles = 24140 m </span>

<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>

<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>

<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>

<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>

<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>

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8 0
3 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

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