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elena55 [62]
3 years ago
10

Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of

Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B. B) three times smaller than the flux of electric field through Gaussian surface B. C) unrelated to the flux of electric field through Gaussian surface B. D) equal to the flux of electric field through Gaussian surface B. E) three times larger than the flux of electric field through Gaussian surface B
Physics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

D) equal to the flux of electric field through the Gaussian surface B.

Explanation:

Flux through S(A) = Flux through S (B ) =  Charge inside/ ∈₀

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Peak wavelength of light coming from a star with a temp of 4300k
PSYCHO15rus [73]

from Wein's law

\frac{k}{ \lambda_{max} }  = T \\ \lambda_{max} =  \frac{k}{T}  \\ \lambda_{max} =  \frac{3.898 \times  {10}^{ - 3} }{4300}   \\  = 9.065 \times  {10}^{ - 7}

3 0
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The blank is the core of an atom​
alexdok [17]

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Step-by-step explanation:

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3 years ago
PART ONE
Lina20 [59]

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass.  Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right

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The y-coordinate of the center of gravity is 47/42 a.

4 0
3 years ago
Sonar is used to determine the speed of an object. A 38.0-kHz signal is sent out, and a 40.0-kHz signal is returned. If the spee
ANEK [815]

Answer:

The velocity is  v  =  8.743 \ m/s

Explanation:

From the question we are told that

    The frequency of the signal sent out  is  f_s  =  38.0 \ kHz  =  38.0 *10^{3} \ Hz

    The frequency of the signal received is  f_r  =  40.0 \ kHz  =  40.0 *10^{3} \ Hz

     The  speed of sound is  v_s  =  341 \ m/s

Generally the frequency of the sound received is  mathematically represented as

         f_r =  f_s  [\frac{v_s  + v}{v_s  - v} ]

where v is the velocity of the object

       =>      40 *10^{3} =  38 *10^{3} *   [\frac{341  + v}{341  - v} ]

       =>      1.05263 = \frac{341+v }{341-v}

       =>   358.94 - 1.05263v  =  341 + v

      =>    17.947 =  2.05263 v

      =>    v  =  8.743 \ m/s

7 0
4 years ago
Which of these black holes exerts the weakest tidal forces on something near its event horizon? a. a 10Msun black hole b. a 100M
Darya [45]

The answer is A. You are welcome.

7 0
3 years ago
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