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3 years ago

The greatest force in momentum will be produced by

Physics

1 answer:

tensa zangetsu [6.8K]3 years ago

4 0

A large force over a large time

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There is a 250-m-high cliff at half dome in yosemite national park in california. suppose a boulder breaks loose from the top of

lorasvet [3.4K]

Part A. For this part, we use two equations for linear motion:

y = y0 + v0 t + 0.5 g t^2 ---> 1

vf = v0 + g t ---> 2

First we solve for t using equation 1: y0 = 0 (initial point at top), y = 250 m, v0 = 0 (at rest)

250 = 0.5 (9.8) t^2

t = 7.143 s

Now we solve for final velocity vf using equation 2:

vf = g t

vf = 9.8 (7.143)

vf = 70 m/s

Part B. First we solve for the time it takes for the sound to reach the tourist.

t(sound) = 250 / 335 = 0.746 s

Therefore the total time would be:

t = 0.746 s + 0.300 s

t = 1.05 s

Hence there is enough time for the tourist to get out before the boulder hits him.

7 0

4 years ago

Read 2 more answers

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material o

zavuch27 [327]

Answer:

Explanation:

7 0

3 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and its value depends on the metal. 

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the frequency has an inverse relation with the wavelength

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago

Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from

Akimi4 [234]

The change in the Gibb's free energy per mole (G) is 1.96 J.

The given parameters:

Density of the ice, ρ = 917 kg/m³
Initial pressure, P₁ = 1.0 bar
Final pressure, P₂ = 2.0 bar
Temperature, T = - 10 C
Mass of water = 18 g

The change in the Gibb's free energy per mole (G) is calculated as follows;

$\Delta G = V(P_2-P_1) \\\\$

where;

V is the volume of the ice

$Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L$

Change in pressure;

$P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm$

The change in the Gibb's free energy per mole (G);

$\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G = 0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J$