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Alika [10]
3 years ago
15

Imagine two free electrons that collide elastically in an acidic solution where one electron was moving and the other electron w

as stationary. When the electrons separate the moving electron now has a velocity of 400 m/s and the stationary electron now has a velocity of 200 m/s. What was the initial kinetic energy of the moving electron
Physics
1 answer:
IgorLugansk [536]3 years ago
4 0

Answer: 9.1 × 10^-26 Joule

Explanation:

Since the collision is elastic. The kinetic energy will be conserved. That is, the sum of kinetic energy before collision will be the same as the sum of the

energy after collision.

Mass of an electron = 9.1 × 10^-31 kg

Given that the velocity of the moving electron = 400 m/s and the stationary electron now has a velocity = 200 m/s. 

K.E = 1/2mv^2

Add the two kinetic energies

1/2mV1^2 + 1/2mV2^2

1/2m( V1^2 + V2^2 )

Since they both have common mass

Substitute m and the two velocities

1/2 × 9.1×10^-31( 400^2 + 200^2)

4.55×10^-31 ( 160000 + 40000 )

4.55×10^-31 × 200000

K.E = 9.1 × 10^-26 Joule

Therefore, the initial kinetic energy of the moving electron is 9.1×10^-26 J

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The force of gravity between Earth and Mars will decrease.

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2 years ago
A platinum resistance thermometer has a resistance of 11.50 ohms at 0 oC and 17.35 ohms at 100 oC. Assuming that the resistance
STatiana [176]

Answer:

34.2^{\circ}C

Explanation:

The resistance increases linearly with the temperature - so we can write:

\Delta R = k \Delta T

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\Delta R is the change in resistance

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\Delta T is the variation of temperature

In the first part of the problem, we have

\Delta R = 17.35 - 11.50 =5.85\Omega

\Delta T = 100 -0 = 100^{\circ}C

So the coefficient of proportionality is

k=\frac{\Delta R}{\Delta T}=\frac{5.85}{100}=0.0585 \Omega ^{\circ}C^{-1}

When the resistance is R=13.50\Omega, the change in resistance with respect to the resistance at zero degrees is

\Delta R' = 13.50-11.50 = 2.00 \Omega

So we can find the change in temperature as:

\Delta T' = \frac{\Delta R}{k}=\frac{2.00}{0.0585}=34.2^{\circ}

So the new temperature is

T_f = T_0 + \Delta T' = 0+34.2 = 34.2^{\circ}C

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