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3 years ago

1)why do we get electric shock while holding a live wire barefooted and not when wearing rubber shoes?

1 answer:

katrin2010 [14]3 years ago

5 0

Electricity is always going to take the path of least resistance to ground. The rubber in your shoes is not a conductor of electricity, therefore you are not completing the circuit and you don't get shocked. Your bare feet, on the other hand ARE conductors of electricity, so when you hold the wire, you complete the circuit and become the path of least resistance to ground... ZAP!

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A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe

LuckyWell [14K]

Answer:

534.9 N

Explanation:

The skier weight is his mass times gravitational acceleration g

W = mg = 103 * 9.8 = 1009.4 N

This weight can be divided into 2 components, one perpendicular and the other parallel to the 32-degree slope. The parallel component would equal to

$Wsin(32^0) = 1009.4sin(32^0) = 534.9 N$

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Strong evidence for the existence of dark matter comes from observations of

adoni [48]

Answer:

NASA's Chandra X-ray Observatory and other telescopes such as the Hubble telescope.

Explanation:

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The above observations have provided the strongest evidence yet that most of the matter in the universe is dark.

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3 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

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