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Iteru [2.4K]
1 year ago
15

On a dry day, just after washing your hair to remove natural oils and drying it thoroughly, run a plastic comb through it. Small

bits of paper will be attracted to the comb. Explain why.
Physics
1 answer:
Hitman42 [59]1 year ago
5 0
When the surface of the comb rubs on your hair, the comb is electrically charged. When the comb comes close to the paper, the charge on the comb causes charge separation on the paper bits. Since paper is neutral, positive and negative charges are equivalent. The charge on the comb charges the area of the bit of paper nearest the comb to the opposite. Thus, the bits of paper become attracted to the comb.
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4. Susan observed that different kinds and amounts of fossils were present in a cliff behind her house. She wondered why changes
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5,10,15,20,25,30, that's how much it should have been
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Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
horsena [70]

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵  kgm² s⁻¹  

Explanation:

a)Given

m = 5500 kg

l = 44 m

Moment of inertia of one blade

I= 1/3 x m l²

where m is mass of the blade

l is length of each blade.

Putting all the required values, moment of inertia of one blade will be

I= 1/3 x 5500 x 44²  

I= 35.49 x 10⁵ kg.m²

Moment of inertia of 3 blades

I= 3 x 35.49 x 10⁵ kg.m²

I= 106.48 x 10⁵ kg.m²

b) Angular momentum 'L' is given by

L =I x ω

where,

I= moment of inertia of turbine i.e  106.48 x 10⁵ kg.m²

ω=angular velocity =2π f

f is frequency of rotation of blade i.e  13 rpm

f = 13 rpm=>= 13 / 60 revolution per second

ω = 2π f =>  2π  x  13 / 60 rad / s

L=I x ω =>106.48 x 10⁵ x   2π  x  13 / 60

  = 144.97 x 10⁵  kgm² s⁻¹    

7 0
1 year ago
Light is propagated as a transverse wave. For this reason, sunglasses, ski goggles and camera lenses can restrict the vibration
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Polerization is the anwser

6 0
2 years ago
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What conclusion can be derived by comparing the central tendencies of the two data sets?
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3 0
2 years ago
A 60-μC charge is held fixed at the origin and a −20-μC charge is held fixed on the x axis at a point x = 1.0 m. If a 10-μC char
Aleksandr [31]

Answer:

Ek =  8,79 [J]

Explanation:

We are going to solve this problem, using  the energy conservation principle

State 1 or initial state (charges at rest t=0)

E₁  = Ek  + U₁

As charge are at rest Ek = 0

And  U₁ has two components

U₁₂   = K * Q₁*Q₂ / 0,4          and    U₃₂  = K*Q₃*Q₂ / 0,6

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,4  ⇒ U₁₂ = 9*60*10*10⁻³/0,4

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,6  ⇒ U₃₂ = - 9*20*10*10⁻³/0,6

U₁₂ = 540*10⁻2/0,4 [J]   ⇒13,5 [J]

U₃₂ = - 180*10⁻² /0,6 [J] ⇒ - 3 [J]

Then   E₁ = E₁₂ + E₃₂    

E₁ = 10,5 [J]

At  the moment of Q₂ passing x = 40 cm  or 0,4 m

E₂ = Ek + U₂

We can calculate the components of U₂ in this new configuration

U₂  =  U₁₂  + U₃₂

U₁₂  = 9*10⁹* 60*10⁻⁶*10*10⁻⁶/0,7   ⇒  U₁₂ = 9*60*10*10⁻³/0,7

U₁₂ = 540*10⁻²/0,7       U₁₂ = 7,71 [J]

U₃₂ =  - 9*10⁹* 20*10⁻⁶*10*10⁻⁶/0,3  ⇒ U₃₂ = -  9*20*10*10⁻³/0,3

U₃₂ = -  9*20*10⁻²/0,3  

U₃₂ = - 6

U₂ = 7,71 -6

U₂ = 1,71 [J]

Then as  

E₂  = Ek + U₂  and  E₂ = E₁

Then

Ek + U₂ = E₁

Ek =  10,5 - U₂    

Ek  = 10,5 - 1,71

Ek =  8,79 [J]

5 0
1 year ago
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