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2 years ago

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The wavelength of an electron whose velocity is 1.7 Ã 104 m/s and whose mass is 9.1 Ã 10-28 g is ________ m. the wavelength of a

n electron whose velocity is 1.7 Ã 104 m/s and whose mass is 9.1 Ã 10-28 g is ________ m. 12 4.3 Ã 10-8 4.3 Ã 10-11 2.3 Ã 10-7 2.3 Ã 107

1 answer:

madreJ [45]2 years ago

4 0

<span>Mass of the electron = 9.1 x 10 ^ -28g = 9.1 x 10 ^ -31kg Velocity of the electron = 1.7 x 10 ^ 4 We have Planck Constant h = 6.626 x 10 ^ -34 Wavelength of the electron w = h/mv w = 6.626 x 10 ^ -34 / ((9.1 x 10 ^ -31)(1.7 x 10 ^ 4)) = 6.626 x 10 ^ -34 / 15.47 x 10 ^ -27 = 0.428312 x 10 ^ -7 = 4.28 x 10 ^ -8 m</span>

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The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. Assume that the rope and pu

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Answer:

$a=2.36\ m/s^2$

T=157.06 N

Explanation:

Given that

Mass of first block = 21.1 kg

Mass of second block = 12.9 kg

First mass is heavier than first that is why mass second first will go downward and mass second will go upward.

Given that pulley and string is mass less that is why both mass will have same acceleration.So lets take their acceleration is 'a'.

So now from force equation

$m_1g-m_2g=(m_1+m_2)a$

21.1 x 9.81 - 12.9 x 9.81 =(21.1+12.9) a

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2 years ago

The half-life of Iodine-131 is 8.0252 days. If 14.2 grams of I-131 is released in Japan and takes 31.8 days to travel across the

MakcuM [25]

Answer:

Explanation:

Half-life problems are modeled as exponential equations. The half-life formula is $P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}$ where $P_o$ is the initial amount, $k$ is the length of the half-life, $t$ is the amount of time that has elapsed since the initial measurement was taken, and $P$ is the amount that remains at time $t$ .

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<u>Deriving the half-life formula</u>

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, $y=a b^{t}$ , where $t$ is still the amount of time, and $y$ is the amount remaining at time $t$ . The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

$y=a b^{t}$

$(14.2)=ab^{(0)}$

$14.2=a *1$

$14.2=a$

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Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

$\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}$

$\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}$

$0.5=b^{8.0252}$

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With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

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$y=0.9108598257$

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