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malfutka [58]
3 years ago
9

How does electricity work in a vacuum

Physics
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

Explanation:

Electricity is a flow of electrons. Electrons can flow across a vacuum. ... They need to leap because the vacuum is a perfect insulator and so there is no medium in which they can flow (like through a metal conductor) so they must aquire all of the energy necessary to cover the distance before they can escape the cathode.

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Un avión tarda en llegar a su destino 12 horas. Si recorrió una distancia de 10.700 kilómetros. Calcular su velocidad y expesarl
elixir [45]

Given that,

Distance, d = 10700 km

Time taken by the airplane to complete the destination, t = 12 hours

We need to find the speed of the airplane. It is equal to the total distance covered divided by total time taken. So,

v=\dfrac{d}{t}\\\\v=\dfrac{10700\ km}{12\ h}\\\\v=891.66\ km/h

We know that,

1 km = 1000 m

1 h = 3600 s

So,

v=891.66\ km/h =\dfrac{891.66\times 1000\ m}{3600\ s}\\\\v=247.68\ m/s

So, the speed of the airplane is 891.66 km/h or 247.68 m/s.

5 0
3 years ago
The diagram shows waves of sound travelling through the air. By which factor is the sound intensity decreased at 3 meters?
SIZIF [17.4K]
The diagram is missing; however, we know that the intensity of a sound wave is inversely proportional to the square of the distance from the source:
I(r)= \frac{1}{r^2}
where I is the intensity and r is the distance from the source.

We can assume for instance that the initial distance from the source is r=1 m, so that we put 
I= \frac{1}{r^2}= \frac{1}{(1)^2}=1
The intensity at r=3 m will be
I= \frac{1}{r^2}= \frac{1}{(3)^2}= \frac{1}{9}
Therefore, the sound intensity has decreased by a factor 1/9.
8 0
3 years ago
Read 2 more answers
Uniformly charged ring with 180 nC/m and radius R= 58 cm. Find the magnitude of the electric field in KN/C at a point P on the a
raketka [301]

Answer:

3.135 kN/C

Explanation:

The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}

Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m

So, E = qz/{4πε₀[√(z² + R²)]³}

E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}

E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}

E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}

E = 6.839 × 10³ Cm²/[1.297 m]³F

E = 6.839 × 10³ Cm²/2.182 m³F

E = 3.135 × 10³ V/m

E = 3.135 × 10³ N/C

E = 3.135 kN/C

3 0
3 years ago
A cubical block of iron 10 cm on each side is floating on mercury in a vessel. (i) What is the height of the block above the mer
Gre4nikov [31]

Answer:

i 5.3 cm ii. 72 cm

Explanation:

i

We know upthrust on iron = weight of mercury displaced

To balance, the weight of iron = weight of mercury displaced . So

ρ₁V₁g = ρ₂V₂g

ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?

V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³

So, the height of iron above the mercury is h = V₂/area of base iron block

= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm

ρ₁V₁g = ρ₂V₂g

ii

ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?

V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ =  7200 cm³

So, the height of column of water is h = V₃/area of base iron block

= 7200 cm³/10² cm² = 72 cm

7 0
3 years ago
The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built?
Sonja [21]
Neglecting friction and air resistance, the first hill must be built 4 times higher than it is now.
3 0
3 years ago
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