Neon signs require about 12,000 V for their operation. Consider a neon-sign transformer that operates off 120- V lines. How many more turns should be on the secondary compared with the primary?

Answers

Answer 1

Answer: I can't give you the actual number of turns, because it's the RATIO
that counts.

However many turns the primary has, the secondary should have
about TEN TIMES that number. Then the transformer will multiply
the primary voltage by 10 ... 120 volts of AC at the primary will
become 1,200 volts of AC at the secondary.

Note that it HAS TObe AC. If the transformer is supplied with DC,
then 120 volts at the primary becomes zero volts at the secondary
and a big cloud of stinky smoke in the room.

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How does the geosphere interact with the hydrosphere

Answers

Plants (biosphere) draw water (hydrosphere) and nutrients from the soil (geosphere) and release water vapor into the atmosphere.

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Answers

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_(f)$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_(cm )^(2)$ + ½ $I_(cm)$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?

Answers

Answer:

the googles are 5.3 m from the edge

Explanation:

Given that

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

$i=arctan(2.2/.90)$

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$1* sin(67.8) = 1.33* sin(r)$

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

A certain lightning bolt moves 50.0 c of charge. how many fundamental units of charge (qe) is this?

Answers

The lightning bolt that moves 50.0 C of charge is converted to qe units. The conversion factor is 1 coulumb is equivalent to 6.24150975·10^18 e. Thus, converting 50 C to e is:

50.0 C * (6.24150975x10^18 e)/1C
equals 3.120754875x10^20 e

Answer: There are $3.125\tims 10^(20)$ number of electrons.

Explanation:

We are given 50 Coulombs of charge and we need to find the number of electrons that can hold this much amount of charge. So, to calculate that we will use the equation:

$nq_e=Q$