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In-s [12.5K]
3 years ago
7

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

Physics
1 answer:
sweet [91]3 years ago
7 0

i will say that 32.2m s2 bt am not sure

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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


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Darryl finds a bottle of what looks like clear water, with dirt settled at the bottom. When he shakes the bottle, the water gets
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Was there any choice of answers so i can help or something u have to figure out
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Which type of wave is more dangerous for humans? Microwaves or x-rays?
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X- rays. It can damage living tissue
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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
3 years ago
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