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bija089 [108]
3 years ago
8

Determine the minimum size of glass tubing that can be used to measure water level. If the capillary rise in the tube does not e

xceed 0.25mm. Take surface tension of water in contact. With air as 0.0735N/m
Physics
1 answer:
Marina CMI [18]3 years ago
8 0

Answer:

Explanation:

The formula to determine the size of a capillary tube is

h = 2•T•Cos θ / r•ρ•g

Where

h = height of liquid level

T = surface tension

r = radius of capillary tube

ρ = density of liquid

θ = angle of contact = 0°

g =acceleration due to gravity=9.81m/s²

The liquid is water then,

ρ = 1000 kg / m³

Given that,

T = 0.0735 N/m

h = 0.25mm = 0.25 × 10^-3m

Then,

r = 2•T•Cos θ / h•ρ•g

r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81

r = 5.99 × 10^-3m

Then, r ≈ 6mm

The radius of the capillary tube is 6mm

So, the minimum size is

Volume = πr²h

Volume = π × 6² × 0.25

V = 2.83 mm³

The minimum size of the capillary tube is 2.83mm³

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Time = (distance) / (speed)

Time = (9.3 x 10^7 miles) x (1609 m/mile) / (3 x 10^8 m/s) = 498.8 seconds .

That would be  <em>8.31 minutes</em>.  
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3 years ago
A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60 to a uniform electric fi
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Answer:

a. 1.75 Nm²/C

b. Yes.

Explanation:

a. Electric Flux is given as:

Φ = E*A*cosθ

Where E = electric flux

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Φ = 14 * 0.25 * cos60

Φ = 1.75 Nm²/C

b. Yes, the shape of the sheet will affect the Flux through it. This is because flux is dependent on area of the surface and the area is dependent on the shape of the surface.

4 0
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Air "breaks down" when the electric field strength reaches 3×106n/c, causing a spark. a parallel-plate capacitor is made from tw
aalyn [17]
Ok, I think this is right but I am not sure:
 Q = ϵ
 0AE
A= π π
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=(8.85x10^-12 C^2/Nm^2)
( π π (0.02m)^2)
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4 0
3 years ago
A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn
Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

AB = 155 ft

Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft

Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft

Using Pythagorean theorem in triangle ADC

AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft

d = distance between the anchor points

distance between the anchor points is given as

d = BD + CD = 70.4 + 93.4\\d = 163.8 ft

5 0
3 years ago
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If I were to transmit a radio wave in our three dimensional world could a fourth dimensional “being” be able to receive it?
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Depends. Are you talking about a mathematical 4th dimension (in which there is infinite dimensions) or some sort of etheral dimension (in which there is no scientific evidence for)

If you mean the first then yes. But it depends how these beings exist. From our understanding we only can theorize shapes in 4-d and if we assume that there is only one universe these "beings" arleady exist and thus any message in 3-d would be sent to them like a shadow ("flat").
If they exist in a alternate "plane" then you would need some method to transverse this plan and if u did, then we would easily be able to communicate, but we would at first sound like a wild animal. They either would ignore us, not understand or perceive us, or they would attempt to send back a signal (essential they are ET's)

IF you mean the second then thats some mystic stuff and its pretty creepy (although a fun read for me :P)
 
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5 0
3 years ago
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