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3 years ago

Helpp T^T

Physics

1 answer:

viva [34]3 years ago

8 0

Answer:

i know this it is actually the second one it develops solar wind. i know this because i am studing this and learning about it

Explanation:

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How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 23.0 cm in diameter to p

lyudmila [28]

Answer:

10573375000

$216.57162\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

r = Distance = $\dfrac{d}{2}=\dfrac{23}{2}=11.5\ cm$

E = Electric field = 1150 N/C

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow q=\dfrac{Er^2}{k}\\\Rightarrow q=\dfrac{1150\times 0.115^2}{8.99\times 10^9}\\\Rightarrow q=1.69174\times 10^{-9}\ C$

Number of electrons is given by

$n=\dfrac{1.69174\times 10^{-9}}{1.6\times 10^{-19}}\\\Rightarrow n=10573375000$

Number of excess electrons is 10573375000

r = 0.115+0.15 = 0.265 m

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 1.69174\times 10^{-9}}{0.265^2}\\\Rightarrow E=216.57162\ N/C$

The electric field is $216.57162\ N/C$

4 0

3 years ago

You place a piece of aluminum at 250.0∘C ∘ C in 9.00 kg k g of liquid water at 20.0∘C ∘ C . None of the water boils, and the fin

Mice21 [21]

Answer:

m₁ = 0.37 kg

Explanation:

According to Law of conservation of energy:

Heat Lost by Aluminum = Heat Gained by Water

m₁C₁ΔT₁ = m₂C₂ΔT₂

where,

m₁ = mass of piece of aluminum = ?

C₁ = specific heat capacity of aluminum = 900 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 250°C - 22°C = 228°C

m₂ = mass of water = 9 kg

C₂ = specific heat capacity of water = 4200 J/kg.°C

ΔT₁ = Change in temperature of aluminum = 22°C - 20°C = 2°C

Therefore,

m₁(900 J/kg.°C)(228 °C) = (9 kg)(4200 J/kg.°C)(2°C)

m₁ = (75600 J)/(205200 J/kg)

5 0

3 years ago

How is work defined in physics

Tcecarenko [31]

Explanation:

Work is the dot product of the force and displacement vectors.

W = F · d

In other words, it is the force times the parallel component of the distance.

W = F d cos θ, where θ is the angle between the force and distance.

3 0

4 years ago

Three identical particles, q1, q2, and q3, each with charge q = 5.00 μC, are placed along a circle of radius r = 2.00 m at angle

wariber [46]

Answer:

11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

Explanation:

$q$ = magnitude of charge on each particle = 5 μC = 5 x 10⁻⁶ C

$r$ = distance of each particle from center of circle = 2 m

$E$ = Magnitude of electric field at the center by each particle

Magnitude of electric field at the center by each particle is given as

$E = \frac{kq}{r^{2} }$

inserting the values

$E = \frac{(9\times10^{9} )(5\times10^{-6})}{2^{2} }\\E = 11.25\times10^{3} NC^{-1}$

From the diagram , we see that being equal and opposite, the electric fields due to charge q₁ and q₃ cancel out.

So net electric field at center is only due to charge q₂ direction towards positive x-direction

$E_{res}$ = Resultant electric field = 11250 N/C

Direction: 0 deg counterclockwise from positive x-axis

7 0

3 years ago

A uniform bar has two small balls glued to its ends. The bar has length L and mass M, while the balls each have mass m and can b

vazorg [7]

Answer:

Explanation:

Length of bar = L

mass of bar = M

mass of each ball = m

Moment of inertia of the bar about its centre perpendicular to its plane is

$I_{1}=\frac{ML^{2}}{12}$

Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is

$I_{2}=2\times m\times \frac{L^{2}}{4}$

$I_{2}=\frac{mL^{2}}{2}$

Total moment of inertia of the system about the centre of the bar perpendicular to its plane is

I = I1 + I2

$I=\frac{ML^{2}}{12}+\frac{mL^{2}}{2}$

$I=\frac{(M +6m)L^{2}}{12}$

8 0

3 years ago