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Orlov [11]
3 years ago
5

A house is 54.0 ft long and 48 ft wide, and has 8.0-ft-high ceilings. What is the volume of the interior of the house in cubic m

eters and cubic centimeters? Answers needs to be in appropriate significant figures.
Physics
1 answer:
pshichka [43]3 years ago
7 0

Explanation:

Length of house, l = 54 ft = 16.45 m

Breadth of house, b = 48 ft = 14.63 m

Height of house, h = 8 ft = 2.43 m

We need to find the volume of the interior of the house. The house is in the form of cuboid. The volume of cuboid is given by :

V=l\times b\times h

V=16.45\ m\times 14.63\ m\times 2.43\ m

V=584.81\ m^3

Since, 1\ m^3=1000000\ cm^3

or the volume of the interior of the house, V=5.84\times 10^8\ cm^3

Hence, this is the required solution.

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A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.
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Answer:

a

 \frac{d \phi_{E}}{dt}  =1.1977 *10^{10} \  V\cdot m/s

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From the question we are told that

  The current is  I =  0.106 \  A

   The length of one side of the square a = 4.60 \  cm = 0.046 \  m

    The separation between the plate is  d = 4.0 mm  = 0.004 \ m

Generally electric flux is mathematically represented as

       \phi_E = \frac{Q}{\epsilon_o}

differentiating both sides with respect to t is  

       \frac{d \phi_{E}}{dt}  = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}

=>     \frac{d \phi_{E}}{dt}  = \frac{1}{\epsilon_o} *I

Here \epsilon_o is the permitivity of free space with value  

        \epsilon _o  =  8.85*10^{-12} C/(V \cdot m)

=>   \frac{d \phi_{E}}{dt}  = \frac{0.106}{8.85*10^{-12}}

=>   \frac{d \phi_{E}}{dt}  =1.1977 *10^{10} \  V\cdot m/s

Generally the displacement current between the plates in A

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=>  I = 0.106 \  A

 

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