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liubo4ka [24]
3 years ago
9

Two copper wires with different diameters are joined end to

Physics
2 answers:
Harman [31]3 years ago
7 0
The current remains same in an end to end (or series) connection ! so, with decrease in cross- sectional area, drift velocity must increase!They experience an "impedance bump" which will affect the total current flow in the circuit.So,current remains constant while th drift velocity increases in areas of less cross-section area..
xxTIMURxx [149]3 years ago
6 0
Decrease because the gauge of the wires are small
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Fifteen identical particles have various speeds: one has a speed of 2.00 m/s, two have speeds of 3.00 m/s, three have speeds of
Evgesh-ka [11]

Answer:

a)  V=7.5m/s

b) rms=8.4m/s

c) Generally the most probable speed is 8m/s as it the most posses by particles being the average

Explanation:

From the question we are told that:

Sample size N=15

  Speed 1 v_1=2m/s\\\\Speed 2 v_2=3m/s\\\\Speed 3 v_2=5m/s\\\\Speed 4 v_4=8m/s\\\\Speed 3 v_5=9m/s\\\\Speed 2 v_6=15m/s\\\\

Generally the equation for Average speed is mathematically given by

 V_{avg}=\frac{\sum(nv)}{N}

Therefore

 V_{avg}=\frac{(2+2(3)+3(5)+4(8)+3(9)+2(15))}{15}

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b)

Generally the equation for RMS speed of the particle is mathematically given by

  rms=\sqrt{\frac{\sum(nv^2)}{N}}

  rms=\sqrt{\frac{2^2+2(3)^2+3(5)^2+4(8)^2+3(9)^2+2(15)^2}{15}}

  rms=\sqrt{69.73}

  rms=8.4m/s

c

Generally the most probable speed is 8m/s as it the most posses by particles being the average

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3 years ago
Which of the following is not an assumption of stage theories?
olya-2409 [2.1K]

Answer:

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Explanation:

8 0
3 years ago
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023 (part 1 of 2) 10.0 points
Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

\Delta KE_{rotational} = \dfrac{1}{2}  \cdot I \cdot \omega ^2 = \dfrac{1}{2}  \cdot m \cdot v ^2

I  = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

\Delta KE_{rotational} = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

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What is the purpose of the article physicists test telepathy in a cheat proof setting?
ryzh [129]
I'm guessing that the purpose of the article is to inform the
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3 years ago
Can someone buy me<br> it will make my day
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Answer:

how much?

Explanation:

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