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ipn [44]
3 years ago
13

A small plastic bead has been charged to -14 nC .

Physics
1 answer:
Studentka2010 [4]3 years ago
3 0

Answer:

The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 5.6 x10^13m/s².

The magnitude of the acceleration of a proton at a distance of 1.5cm from the bead is 9.8 x10^16m/s².

Explanation:

Newton's second law states that the total sum of the force acting on a particle in motion is equal to the mass of the particle times the acceleration due to the force. So the electric force between the bead and proton is equal to ma. That is,

Fe = kq1*q2/r² = m*a

The proton had a charge of +1.6x10^-19C and a mass of 1.67×10^-27kg

By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 5.6×10^13m/s²

The proton had a charge of -1.6x10^-19C and a mass of 9.10×10^-31kg

By substituting these values into the equivalent for a we have that the acceleration of the proton at a distance of 1.5cm form the bead is 9.8×10^16m/s²

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Answer:

Thrust = 200 N

Explanation:

The engine thrust can be found by using the following formula:

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v = velocity of ejected gases = 4000 m/s

Therefore, using the given values in the equation, we get:

Thrust = (0.05\ kg/s)(4000\ m/s)

<u>Thrust = 200 N</u>

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An object blocks the path of a wave causing the wave to change direction is called what
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Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains
OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

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3 years ago
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