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Degger [83]
3 years ago
10

The charge on a parallel plate capacitor is 7.68 C, and the plate is connected to a 9.0 V battery. What is the capacitance of th

e capacitor?
A. 8.5x10^-7 pF
B. 1.1x10^-6 pF
C. 8.5x10^5 pF
D. 6.9x10^7 pF
Physics
1 answer:
Natali5045456 [20]3 years ago
8 0

Charge on a capacitor = (capacitance) x (voltage)

7.68 Coulomb = (capacitance) x (9.0 V)

Capacitance = (7.68 Coul) / (9.0 V)

<em>Capacitance = 0.83 Farad</em>

<em></em>

Apparently, the " 10^something " after the 7.68 C got lost.

If the charge is actually 7.68 <u>micro</u>coulombs, then the answer is<em> (A) .</em>

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A plane has an airspeed of 142 m/s. A 30.0 m/s wind is blowing southward at the same time as the plane is flying. What must be t
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Answer:

\theta=12.19^{\circ}

Explanation:

Given that

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The speed of the air ,u = 30 m/s

Lets take angle make by airplane from east direction towards north direction is θ .

Now by using diagram ,we can say that

sin\theta =\dfrac{u}{v}

Now by putting the values in the above equation we get

sin\theta =\dfrac{30}{142}

sin\theta=0.21

\theta=12.19^{\circ}

Therefore the angle will be 12.19° .

 

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You need a 450 microgram sample of gold, but you only have a mass balance that measures in decigrams. Convert the amount of gold
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The amount of gold in decigrams if 450 micrograms is needed is 4.5 × 10-³ decigrams.

<h3>How to convert micrograms to decigrams?</h3>

According to this question, 450 micrograms of a sample of gold is needed but we only have a mass balance that measures in decigrams.

This means that we are to convert the amount of gold you need to decigrams by comparing the exponents.

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A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax
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Answer:

From the question we are told that

  The length of the rod is  L_o

    The  speed is  v  

     The angle made by the rod is  \theta

     

Generally the x-component of the rod's length is  

     L_x =  L_o cos (\theta )

Generally the length of the rod along the x-axis  as seen by the observer, is mathematically defined by the theory of  relativity as

       L_xo  =  L_x  \sqrt{1  - \frac{v^2}{c^2} }

=>     L_xo  =  [L_o cos (\theta )]  \sqrt{1  - \frac{v^2}{c^2} }

Generally the y-component of the rods length  is mathematically represented as

      L_y  =  L_o  sin (\theta)

Generally the length of the rod along the y-axis  as seen by the observer, is   also equivalent to the actual  length of the rod along the y-axis i.e L_y

    Generally the resultant length of the rod as seen by the observer is mathematically represented as

     L_r  =  \sqrt{ L_{xo} ^2 + L_y^2}

=>  L_r  = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}

=>  L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}

=>   L_r  = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}

=> L_r =  \sqrt{L_o^2 * cos^2(\theta)  [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}

=> L_r  =  \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }

=> L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

Hence the length of the rod as measured by a stationary observer is

       L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }

   Generally the angle made is mathematically represented

tan(\theta) =  \frac{L_y}{L_x}

=>  tan {\theta } =  \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }

=> tan(\theta ) =  \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }

Explanation:

     

     

       

7 0
3 years ago
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