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fgiga [73]
3 years ago
11

At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re

st at one of these locations passes through the other location. From which location was the proton released? a. B b. А What is its speed when it passes through the other location? speed: m /s Repeat the same question, but this time for an electron.
From which location was the electron released? a. А b. B What is its speed when it passes through the other location? speed: m/s TOOLS Y10
Physics
1 answer:
densk [106]3 years ago
4 0

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

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Answer: The elimination of seasonal variations

Explanation:

Since the cosmic catastrophic event which occurred led to the tilt of the Earth's axis relative to the plane of orbit to increase from 23.5° to 90°, the most obvious effect of this change would be the elimination of seasonal variations.

It should be noted that seasonal variation refers to the variation in a time series that's within a year which is repeated. The cause of seasonal variation can include rainfall, temperature, etc.

7 0
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One isotope of a metallic element has the mass number 65 and 35 neutrons in the nucleus. the cation derived from the isotope has
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The mass number of an isotope is the sum of the numbers of protons and the numbers of the neutron. Given that the mass number is 65 and the number of neutrons is 35, the number of protons is 30. The atom is then Zinc (Zn). The charge is equal to +2, as it lacks 2 more electrons. 
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2 m3 of an ideal gas are compressed from 100 kPa to 200 kPa. As a result of the process, the internal energy of the gas increase
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Answer:

work done is -150 kJ

Explanation:

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pressure p1 = 100 kPa

pressure p2 = 200 kPa

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solution

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3 0
3 years ago
A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

  • \Delta h is the change in the ball's height.
  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

t \approx 0.440315\;\text{s}.

What's the initial horizontal velocity of the ball?

  • Horizontal displacement of the ball: \Delta x = 0.352\;\text{m};
  • Time taken: \Delta t = 0.440315\;\text{s}

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0
3 years ago
I need help, please answer
Burka [1]

This being a perfect collision means no energy is lost during the collision. Because this question asks for speed and not velocity, the speed will be the same because the final energy is the same. The speed after the collision would therefore be 1.27 m/s.

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