A charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 36 N/C. The electric field 4 cm from the wire is:

Answers

Answer 1

Answer:

New electric field = 18 N/C

Explanation:

Given:

Length (E1) = 2 cm

New length (E2) = 4 cm

Electric field =  36 N/C

Find:

New electric field

Computation:

New electric field = 36 [2 / 4]

New electric field = 36 [1/2]

New electric field = 18 N/C


Related Questions

Input resistance of a FET is very high due to A) forward-biased junctions have high impedance B) gate-source junction is reverse-biased C) drain-source junction is reverse-biased D) none of the above

Answers

Answer:

B) gate-source junction is reverse-biased

Explanation:

FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow"   in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".

In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".

Consider atmospheric air at 25 C and a velocity of 25 m/s flowing over both surfaces of a 1-m-long flat plate that is maintained at 125 C. Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to 105 , 5 105 , and 106 .

Answers

Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

 

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

After clamping a buret to a ring stand, you notice that the set-up is tippy and unstable. What should you do to stabilize the set-up

Answers

Answer:

Move the buret clamp to a ring stand with a larger base.

Explanation:

A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.

The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.

The clamp is used to hold the burette in place.

If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.

The larger base provides a better center of gravity and stabilises the setup

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