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tino4ka555 [31]
3 years ago
15

If one branch of a 120-v power lines is protected by a 20-A fuse, will the fuse carry an 8-Ώ load

Physics
1 answer:
Mrrafil [7]3 years ago
7 0

Answer:

No I won't.

It will carry 6ohm load.

Explanation:

It obeys ohms law therefore V=IR

120=20R

R=120/20

R= 6 ohms

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The definition of density is

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Divide each side by 'density' :                         Volume = (mass) / (density)
 

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E) Thermal energy is released during
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Answer:

e) True, f) False

Explanation:

e) Let consider a close system, that is, a system with no mass interactions with surroundings. Then, we get the following expression by the First Law of Thermodynamics:

Q_{net,in} - W_{net, out} = \Delta U (1)

Where:

Q_{net, in} - Net input heat, measured in joules.

W_{net, out} - Net output work, measured in joules.

\Delta U - Change in thermal energy, measured in joules.

Please notice that work comprises all kind of work (i.e. mechanical, electric, magnetic), whereas heat comprises all heat interactions including chemical and radioactive phenomena.

If thermal energy is released, then \Delta U < 0, which is caused by three scenarios:

(i) Q_{net,in} < 0, W_{net, out} < 0, |Q_{net,in}|>|W_{net,out}|

(ii) Q_{net, in} > 0, W_{net,out} > 0, |Q_{net,in}|

(iii) Q_{net,in}< 0, W_{net, out}>0

In the case Q_{net,in} > 0, W_{net, out}, the thermal energy of the system is increased. Therefore, thermal energy is released during some energy conversions. Answer: True

f) A liquid solidifies when temperature goes below point of fusion, meaning a realease of heat with no work interactions. That is:

Q_{net, in} = \Delta U, Q_{net, in} < 0 (2)

If Q_{net, in} < 0, then  \Delta U < 0. Then, if a liquid absorbs heat energy, then thermal energy is increase and the liquid does not solidifies. Answer: False.

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2 years ago
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2 years ago
To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d
Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by k  =  60 cm  =  0.6 \  m

    The frequency of both speakers is f =  700 \  Hz

Generally the distance of the listener to the first speaker is mathematically represented as

       L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}

       L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}

        L_1  =   5.39 \  m

Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

Here \lambda is the wavelength which is mathematically represented as

         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

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