A toy rocket fires its engines to launch it straight up from rest. After a

A train is approaching a signal tower at a speed of 40m/s. The train engineer sounds the 1000-Hz whistle, while a switchman in t

stealth61 [152]

v = speed of the source of sound or the train towards the listener or switchman = 40 m/s

V = actual speed of sound = 340 m/s

f = actual frequency of sound as emitted from source or the train = 1000 Hz

f' = frequency as observed by the listener or by switchman = ?

Using Doppler's law , frequency observed by a listener from a source moving towards it is given as

f' = V f /(V - v)

inserting the values

f' = 340 x 1000 /(340 - 40)

f' = 340 x 1000/300

3 0

3 years ago

Darryl finds a bottle of what looks like clear water, with dirt settled at the bottom. When he shakes the bottle, the water gets

Alexus [3.1K]

Was there any choice of answers so i can help or something u have to figure out

7 0

3 years ago

Which type of wave is more dangerous for humans? Microwaves or x-rays?

Dvinal [7]

X- rays. It can damage living tissue

3 0

3 years ago

Read 2 more answers

I need help someone help me

GaryK [48]

what do you need , i mean your help , let me see if i can help

5 0

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

3 years ago