A proton moves through an electric potential created by a number of source charges. Its speed is 2.5X105 m/s at a point where the potential is 1500 V. What will be the proton

(6.67 × 10⁵) m/s

Explanation:

What will be the proton's speed a short time later when it reaches a point where the potential is -500V.

According to the law of comservation of energy, the sum of potential and kinetic energy of the proton at any point in time is always constant.

P.E₁ + K.E₁ = P.E₂ + K.E₂

P.E₁ = U₁ = qV₁

q = charge on the proton = (1.602 × 10⁻¹⁹) C

V₁ = 1500 V

P.E₁ = U₁ = (1.602 × 10⁻¹⁹) × (1500) = (2.403 × 10⁻¹⁶) J

K.E₁ = (1/2)mv₁²

m = mass of a proton = (1.673 × 10⁻²⁷) kg

v₁ = (2.5 × 10⁵) m/s

K.E₁ = (1/2)(1.673×10⁻²⁷)(2.5 × 10⁵)² = (5.23 × 10⁻¹⁷) J

P.E₂ = U₂ = qV₂

q = charge on the proton = (1.602 × 10⁻¹⁹) C

V₂ = -500 V

P.E₂ = U₂ = (1.602 × 10⁻¹⁹) × (-500) = (-8.01 × 10⁻¹⁷) J

K.E₂ = ?

P.E₁ + K.E₁ = P.E₂ + K.E₂

(2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) = (-8.01 × 10⁻¹⁷) + K.E₂

K.E₂ = (2.403 × 10⁻¹⁶) + (5.23 × 10⁻¹⁷) + (8.01 × 10⁻¹⁷) = (3.727 × 10⁻¹⁶) J

K.E₂ = (1/2)mv₂²

m = (1.673 × 10⁻²⁷) kg

v₂ = ?

(3.727 × 10⁻¹⁶) = (1/2)(1.673 × 10⁻²⁷)v₂²

v₂² = (4.455 × 10¹¹)

v₂ = 667,457.863839 m/s = (6.67 × 10⁵) m/s

Hope this Helps

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

which piece of furniture will have the most inertia and give the furniture movers the most difficulty in moving?  

answer

a 75kg dining table