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3 years ago

a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.

Physics

1 answer:

Allushta [10]3 years ago

4 0

Answer:

=3 metre per second ^2

Explanation:

Formula for acceleration is

V-U÷T

In the given information

V=16

U=4

T=4

Acceleration =16-4/4

=3 metre per second ^2

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Sarah bikes 15 miles in 70 minutes. Which is her average speed per minute?

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15/70 =3/14 miles per minute = 0.21 miles a minute. One mile in about 5 minutes. Sarah is shifting, I think (faster than me)

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4 years ago

A carton of 12 rechargable batteries contains one that is defective. In how many ways can the inspector choose 3 of the batterie

atroni [7]

There are 2 defective batteries and 10 good batteries.
a) none of the defective batteries (10C3) = 120
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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

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

7 0

4 years ago

If light of wavelength 700 nm strikes such a photocathode, what will be the maximum kinetic energy, in eV , of the emitted elect

Oksana_A [137]

If the light of wavelength 700 nm strikes such a photocathode the maximum kinetic energy, in eV, of the emitted electrons is 0.558 eV.

so - $KE_{max} = hc/lembda} work

threshold when KE = 0

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What is photocathode?

A photocathode electrolyte interface can be used in a photoelectrolysis cell as the primary light-harvesting junction (in conjunction with an appropriate electrochemical anode) or as an optically complementary photoactive half-cell in a tandem photoelectrode photoelectrolysis cell (Hamnett, 1982; Kocha et al, 1994).
In the case of the former, the electrode should ideally harvest photon energy across the majority of the solar spectrum in order to achieve the highest energy conversion efficiency possible.
In the latter case, however, the photocathode may only be active in a specific band of the solar spectrum in order to generate a cathodic photocurrent sufficient to match the current generated in the photoanodic half-cell.

To learn more about Photocathode from the given link:

brainly.com/question/9861585

#SPJ4

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2 years ago

A collector of rare items is negotiating with a museum. He wants to sell a spear he claims to have been

viva [34]

Answer:

I'm not sure if you maybe forgot to add in some information but I can give you the equation you will need to solve this..! (just without the bit of infomration you might need)

$2=4(\frac{1}{2})^{\frac{1100}{t}}$

In this case you'll be solving for t and I can help you with your first step, you'll divide 2 by 4, getting you to:

$\frac{1}{2} =\frac{1}{2}^{\frac{1100}{t}$

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3 years ago

Read 2 more answers

a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.​

a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.