A train is traveling at 50km/h average .what is the displacement of the train per second?

The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of

vodomira [7]

Answer:

The new Coulomb force is q₁q₂/9πε₀r²

Explanation

The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².

Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.

Since we have air between q₂ and q₃, the coulomb force between them is

F' = q₂q₃/4πε₀d²

= q₂(q₁/κ)/4πε₀(r/2)²

= 4q₂q₁/κ4πε₀r²

= 4/κ(q₂q₁/4πε₀r²)

= 4/9 × (q₂q₁/4πε₀r²)

= q₁q₂/9πε₀r²

So, the new Coulomb force is q₁q₂/9πε₀r²

3 0

4 years ago

Nevine wants to improve her JavaScript program's efficiency and scalability by defining her own processes, or functions. Why are

alex41 [277]

Because they perform specific tasks repeatedly throughout your program, as needed

3 0

4 years ago

A bus contains a 1440 kg flywheel (a disk that has a 0.63 m radius) and has a total mass of 10200 kg. Calculate the angular velo

CaHeK987 [17]

Answer: $\omega =93.51 rad/s$

Explanation:

Given

mass of Flywheel $m_1=1440 kg$

mass of bus $m_b=10200 kg$

radius of Flywheel $r=0.63 m$

final speed of bus $v=21 m/s$

Conserving Energy i.e.

0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus

Let $\omega$ be the angular velocity of Flywheel

$0.9\cdot \frac{I\omega ^2}{2}=\frac{m_bv^2}{2}$

$I=moment\ of\ Inertia =mr^2=1440\cdot 0.63^2=571.536 kg-m^2$

$0.9\cdot \frac{571.536\cdot \omega ^2}{2}=\frac{10200\cdot 21^2}{2}$

$\omega ^2=21^2\times \frac{10200}{0.9\times 571.536}$

$\omega =21\times 4.45=93.51 rad/s$

8 0

3 years ago

Metals combine easily with nonmetals. true or false

Ad libitum [116K]

The answer for that is True.

4 0

3 years ago

Please help me people

saveliy_v [14]

Answer :

(1) The density of asphalt is, $1200kg/m^3$

(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.

(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

Explanation :

Part 1 :

As we are given:

Mass of block = 90 kg

Volume of block = $0.075m^3$

Formula used :

$\text{Density of block}=\frac{\text{Mass of block}}{\text{Volume of block}}$

Now put all the given values in this formula, we get:

$\text{Density of block}=\frac{90kg}{0.075m^3}=1200kg/m^3$

Thus, the density of asphalt is, $1200kg/m^3$

Part 2(a) :

As we are given that:

Length of aluminium sheet = 35 cm

Width of aluminium sheet = 11 dm

Thickness of aluminium sheet = 15 mm

Now we have to convert these dimensions into meters.

Conversions used:

1 cm = 0.01 m

1 dm = 0.1 m

1 mm = 0.001 m

Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m

Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m

Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m

Part 2(b) :

First we have to calculate the volume of aluminium sheet.

Volume of aluminum sheet (cuboid) = Length × Width × Thickness

Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m

Volume of aluminum sheet (cuboid) = 0.005775 m³

Now we have to calculate the mass of aluminium sheet.

$\text{Density of aluminium}=\frac{\text{Mass of aluminium}}{\text{Volume of aluminium}}$

$2700kg/m^3=\frac{\text{Mass of aluminium}}{0.005775m^3}$

$\text{Mass of aluminium}=15.59kg$

Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.

7 0

3 years ago

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A train is traveling at 50km/h average .what is the displacement of the train per second?​

A train is traveling at 50km/h average .what is the displacement of the train per second?