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3 years ago

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An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength

of 880 nm. What is the potential difference though which this electron was accelerated

1 answer:

Vladimir [108]3 years ago

4 0

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

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Lerok [7]

Answer:

Option A

Explanation:

The Equation represents the displacement of the object which is represented by x

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so, $x_0$ means when time is zero so we replace t with zero in the equation,

$x_0=(0)-(0)^2\\x_0=0$

now for v which is velocity we need to differentiate the function as the formula for velocity is rate of change of displacement over time so we derivate the equation once and get,

$v=1-2t\\$

now for $v_0$ we insert t = 0 and get

$v_0=1-2(0)\\v_0=1$

now for a which is acceleration the formula of acceleration is rate of change of velocity over time, so we differentiate the the equation of v(velocity) once or the equation of x(displacement) twice so now we get,

$a=-2$

so Option A is your answer.

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3 years ago

Read 2 more answers

Calculate the standard electrode potential difference (e°) of the daniell cell (at 1 bar) if temperature is 473.15 k.

anzhelika [568]

Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.

Solution:

1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
$E^0=E_{cat}^0-E_{an}^0$
where $E_{cat}^0$ is the standard potential at the cathode, while $E_{an}^0$ is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: $E_{Cu}^0=+0.34 V$ , while at the anode we have zinc: $E_{Zn}^0=-0.76 V$ . Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
$E^0=+0.34 V-(-0.76 V)=+1.1 V$

2) To calculate $E^0$ at any temperature T, we should use Nerst equation:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}$
where
$R=8.31 J/(K mol)$
$T=473.15 K$ is the temperature in our problem
$z=2$ is the number of electrons transferred in the cell's reaction
$F=9.65\cdot 10^4 C/mol$ is the Faraday's constant
$[Zn]$ and $[Cu]$ are the molar concentrations of zinc and in copper, and in our problem we have $[Zn]=10[Cu]$ .
Using all these data inside the equation, and using $E^0=+1.1 V$ , in the end we find:
$E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V$

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3 years ago

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pav-90 [236]

Answer: This is False.

Explanation:

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3 years ago

shepuryov [24]

Answer:

a) I = 464 kg m², b) K = 631 .6 J, c) v = 8.25 m / s

Explanation:

a) the moment of inertia of point particles is

I = ∑ m_i r_i²

in this case

I = 8 5² + 3 (-2) ² + 7 (-6) ²

I = 464 kg m²

b) The kinetic energy is

K = ½ I w²

K = ½ 464 1.65²

K = 631 .6 J

c) linear and angular velocity are related

v = w r

v = 1.65 5

v = 8.25 m / s

8 0

3 years ago