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erastovalidia [21]
3 years ago
10

As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t

he other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again?As in the video, we apply a charge + to the half-shell that carries the electroscope. This time, we also apply a charge – to the other half-shell. When we bring the two halves together, we observe that the electroscope discharges, just as in the video. What does the electroscope needle do when you separate the two half-shells again?It deflects more than it did at the end of the video.It does not deflect at all.It deflects the same amount as at end of the video.It deflects less than it did at the end of the video.
Physics
2 answers:
frez [133]3 years ago
8 0

Answer:It does not deflect at all.

Explanation:

Andreyy893 years ago
4 0

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

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A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

       Q = 0.0192 \  m^3 /s

6 0
3 years ago
The electrical resistance of an element in a platinum resistance thermometer at 100°c, 0°c and room temperature are 75.00Ω, 63.0
Dominik [7]

Answer:

16.6 °C

Explanation:

From the question given above, the following data were obtained:

Temperature at upper fixed point (Tᵤ) = 100 °C

Resistance at upper fixed point (Rᵤ) = 75 Ω

Temperature at lower fixed point (Tₗ) = 0 °C

Resistance at lower fixed point (Rₗ) = 63.00Ω

Resistance at room temperature (R) = 64.992 Ω

Room temperature (T) =?

T – Tₗ / Tᵤ – Tₗ = R – Rₗ / Rᵤ – Rₗ

T – 0 / 100 – 0 = 64.992 – 63  / 75 – 63

T / 100 = 1.992 / 12

Cross multiply

T × 12 = 100 × 1.992

T × 12 = 199.2

Divide both side by 12

T = 199.2 / 12

T = 16.6 °C

Thus, the room temperature is 16.6 °C

6 0
2 years ago
PLS ANSWERR THIS QUESTION FOR ME !!
MA_775_DIABLO [31]

Explanation:

20 joule is your answer

Answer:

here

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acceleration due to gravity a =10m/s²

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3 0
3 years ago
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Firdavs [7]

Answer:

speed of the bullet before it hit the block is 200 m/s

Explanation:

given data

mass of block m1 = 1.2 kg

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to find out

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solution

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put all value in equation 1

m1 × v1 + m2 × v2 = M × V

1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8

solve it we get

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so speed of the bullet before it hit the block is 200 m/s

8 0
3 years ago
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crimeas [40]

Answer:

position 3

Explanation: HOPE IT HELPED

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