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disa [49]
3 years ago
14

How does charging by conduction occur?

Physics
2 answers:
tia_tia [17]3 years ago
6 0
Hello!
Charging by conduction involves the contact of a charged object to a neutral object. Suppose that a positively charged aluminum plate is touched to a neutral metal sphere. The neutral metal sphere becomes charged as the result of being contacted by the charged aluminum plate. 
Hope this helped!
Nadusha1986 [10]3 years ago
6 0

Answer:

i believe it is a charged object touches a neutral object. please let me know if I'm wrong

Explanation:

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The length of the adult mosquito is typically between 3.0 mm and 6.0 mm. The smallest known mosquitoes are around 2.5 mm, and th
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0.74in

Explanation:

Given parameters:

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Smallest mosquito = 2.5mm

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Unknown:

Length of largest known mosquito in inches = ?

Solution;

The length is the longest dimension. It is how long a body is.

 The problem here is converting from mm to inches;

 

The length of the longest mosquito which is the largest is 19mm

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  19mm = 19mm x \frac{0.039in}{1mm} = 0.74in

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Which of the following is the best definition of lifestyle activity?
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A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and
alexira [117]

Answer:

A) \omega_f=17.503\ rad.s^{-1}

B) t=55.6822\ s

C) \theta=1312\ rad

Explanation:

Given:

  • mass of flywheel, m=40\ kg
  • diameter of flywheel, d=0.72\ m
  • rotational speed of flywheel, N_i=450\ rpm \Rightarrow \omega_i=\frac{450\times 2\pi}{60} =15\pi\ rad.s^{-1}
  • duration for which the power is off, t_0=35\ s
  • no. of revolutions made during the power is off, \theta=180\times 2\pi=360\pi\ rad

<u>Using equation of motion:</u>

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

360\pi=15\pi\times 35+\frac{1}{2} \times \alpha\times35^2

\alpha=-0.8463\ rad.s^{-2}

Negative sign denotes deceleration.

A)

Now using the equation:

\omega_f=\omega_i+\alpha.t

\omega_f=15\pi-0.8463\times 35

\omega_f=17.503\ rad.s^{-1} is the angular velocity of the flywheel when the power comes back.

B)

Here:

\omega_f=0\ rad.s^{-1}

Now using the equation:

\omega_f=\omega_i+\alpha.t

0=15\pi-0.8463\times t

t=55.6822\ s is the time after which the flywheel stops.

C)

Using the equation of motion:

\theta=\omega_i.t+\frac{1}{2} \alpha.t^2

\theta=15\pi\times 55.68225-0.5\times 0.8463\times 55.68225^2

\theta=1312\ rad revolutions are made before stopping.

3 0
3 years ago
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