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1 year ago

What force pulls a falling apple down to the ground?

2 answers:

8 0

Gravity D

Obviously, under the influence of gravity, the apple begins to “fall” downwards. We can even see that the speed of the apple, which at each point is given by the change of r divided by the change of t, is zero at the beginning (t=0) and then increases while the apple moves towards the earth

iren2701 [21]1 year ago

5 0

Answer:

Explanation:

Gravitational force is a type of force that pulls objects to the Earth's surface.

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(a) Find the energy of the ground state (n = 1) and the first two excited states of an electron in a one-dimensional box of leng

const2013 [10]

(a) $3.77\cdot 10^5 MeV, 1.51\cdot 10^6 MeV, 3.39\cdot 10^3 GeV$

The energy levels of an electron in a box are given by

$E_n = \frac{n^2 h^2}{8mL^2}$

where

n is the energy level

$h=6.63\cdot 10^{-34}Js$ is the Planck constant

$m=9.11\cdot 10^{-31}kg$ is the mass of the electron

$L=1.0\cdot 10^{-15} m$ is the size of the box

Substituting n=1, we find the energy of the ground state:

$E_1 = \frac{1^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=6.03\cdot 10^{-8}J$

Converting into MeV,

$E_1 = \frac{6.03\cdot 10^{-8} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =3.77\cdot 10^5 MeV$

Substituting n=2, we find the energy of the first excited state:

$E_2 = \frac{2^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=2.41\cdot 10^{-7}J$

Converting into MeV,

$E_2 = \frac{2.41\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-6} MeV/eV =1.51\cdot 10^6 MeV$

Substituting n=3, we find the energy of the second excited state:

$E_3 = \frac{3^2 (6.63\cdot 10^{-34}^2}{8(9.11\cdot 10^{-31}(1.0\cdot 10^{-15})^2}=5.43\cdot 10^{-7}J$

Converting into GeV,

$E_3 = \frac{5.43\cdot 10^{-7} J}{1.6\cdot 10^{-19} J/eV}\cdot 10^{-9} GeV/eV =3.39\cdot 10^3 GeV$

(b) $1.10 \cdot 10^{-18} m$

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

$E=E_2 - E_1 = 2.41\cdot 10^{-7}J - 6.03\cdot 10^{-8} J=1.81\cdot 10^{-7} J$

And the energy of the electromagnetic radiation is

$E=\frac{hc}{\lambda}$

where c is the speed of light; so, re-arranging the formula, we find the wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.81\cdot 10^{-7}J}=1.10 \cdot 10^{-18} m$

(c) $6.59 \cdot 10^{-19} m$

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

$E=E_3 - E_2 = 5.43\cdot 10^{-7} J - 2.41\cdot 10^{-7}J =3.02\cdot 10^{-7} J$

Using the same formula as before, we find the corresponding wavelength:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.02\cdot 10^{-7}J}=6.59 \cdot 10^{-19} m$

(d) $4.12 \cdot 10^{-19} m$

The energy of the emitted radiation is equal to the energy difference between the two levels, so:

$E=E_3 - E_1 = 5.43\cdot 10^{-7} J - 6.03\cdot 10^{-8}J =4.83\cdot 10^{-7} J$

Using the same formula as before, we find:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4.83\cdot 10^{-7}J}=4.12 \cdot 10^{-19} m$

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3 years ago

Answer me fast. Find velocity in Acceleration Time Graph.

Ipatiy [6.2K]

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2 years ago

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Which component of the atom has the least mass?

mrs_skeptik [129]

B) Electrons have the least amount of mass

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3 years ago

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Which graph shows uniform motion.

bagirrra123 [75]

Answer:

1. shows uniform motion

Explanation:

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3 years ago

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A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago