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3 years ago

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A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J

. Calculate the kinetic energy and speed of the ejected electron.

1 answer:

Lyrx [107]3 years ago

7 0

Explanation:

Given that,

Wavelength of the light, $\lambda=4170\ A=4170\times 10^{-10}\ m$

Work function of sodium, $W_o=4.41\times 10^{-19}\ J$

The kinetic energy of the ejected electron in terms of work function is given by :

$KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J$

The formula of kinetic energy is given by :

$KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s$

Hence, this is the required solution.

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A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.

babunello [35]

Answer:

True

Explanation:

Pressure is defined as:

$p=\frac{F}{A}$

where

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Therefore, pressure is inversely proportional to the area of the surface:

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this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

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3 years ago

A watermelon is dropped from the top of a 80m tall building. We want to find the velocity of the watermelon after it falls for 1

White raven [17]

Answer:

$v_f = v_i + at$

$v_f = 13.23 m/s$

Explanation:

Height Of the watermelon when it is dropped is given as

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time of fall under gravity

$t = 1.35 s$

now if water melon start from rest then we have

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acceleration due to gravity for watermelon

$a = 9.81 m/s^2$

now we need to find the final speed of watermelon

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so we will have

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levacccp [35]

Answer:

$\alpha = 6.431\,\frac{rad}{s^{2}}$

Explanation:

The pulley is modelled by the Newton's Laws, whose equation of equilibrium is:

$\Sigma M = T \cdot R = \frac{1}{2}\cdot M \cdot R^{2}\cdot \alpha$

Given that tension is equal to the weight of the bucket, the angular acceleration experimented by the pulley is:

$T = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$m_{b}\cdot g = \frac{1}{2}\cdot M \cdot R \cdot \alpha$

$\alpha = \frac{2\cdot m_{b}\cdot g}{M\cdot R}$

$\alpha = \frac{2\cdot (1.53\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{(7.07\,kg)\cdot (0.66\,m)}$

$\alpha = 6.431\,\frac{rad}{s^{2}}$

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3 years ago

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Anarel [89]

::Answer::

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3 years ago

What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as

Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

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2 years ago