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4 years ago

Guys I need help with my homework please !!!!

Physics

1 answer:

Sergio [31]4 years ago

6 0

Answer:

(a) To draw water from a well we have to pull at the rope.

(b) A charged body attracts an uncharged body towards it.

(d) The north pole of a magnet repels the north pole of another magnet.

Explanation:

Just trust me

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True or false that humans emit light?

krok68 [10]

I'm pretty sure it's true. We glow with "Bioluminescence".

4 0

3 years ago

The process by which engine fuels burn is called ____.

Paul [167]

Answer:

The answer is Combustion

8 0

4 years ago

1. An electron in an atom absorbs a photon with an energy of 2.38 eV and jumps from the n = 2 to n = 4 energy level in the atom.

oksian1 [2.3K]

1. $1.0\cdot 10^{-6}m$

First of all, let's convert the energy of the absorbed photon into Joules:

$E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J$

The energy of the photon can be rewritten as:

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

where

h is the Planck constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m$

2. 1.24 eV

In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength

$\lambda=1.66\cdot 10^{-6}m$

So the energy of the emitted photon is given by the formula used previously:

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

and using

$\lambda=1.66\cdot 10^{-6}m$

we find

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J$

converting into electronvolts,

$E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV$

EDIT: an issue in Brainly does not allow me to add the last 2 parts of the solution - I have added them as an attachment to this post, check the figure in attachment.

7 0

3 years ago

A boat with a mass of 1000 kg drifts with the current down a straight section of river parallel to the +x+x axis with a speed of

IrinaK [193]

Answer:

Explanation:

Our values are defined by,

$m=1000kg\\v = 2m/s\\a = 2.7 \angle 45\°$

We can express also in a vectorial way, then

$v=2\hat{i} \rightarrow$ no values in $\hat{j}$

Acceleration as follow,

$a= 2.7cos45 \hat{i} + 2.7 sin45 \hat{j}$

We know that velocity is given by,

$V(t) = v_0+at \rightarrow v_0 = 2$

We need to calculate for t=3, then

$v(3) = 2\hat{i} +(2.7cos45 \hat{i} + 2.7 sin45 \hat{j})*3$

$v(3) = (2\hat{i}+3*2.7cos45 \hat{i})+(3*2.7 sin45 \hat{j})$

$v(3) = (2+3*2.7cos45)\hat{i}+(5.7 sin45 )\hat{j}$

$v(3) = 7.7\hat{i}+5.7\hat{j}$

Our mass is 1000Kg, so the momentum is

$P = mv$

$P= 1000(7.7\hat{i}+5.7\hat{j})$

$P=7700\hat{i}+5700\hat{j}$

7 0

4 years ago

2. Un ciclista que está en reposo comienza a pedalear hasta alcanzar los 16.6km/h en 6 minutos. Calcular la distancia total que

jek_recluse [69]

Answer:

$s = 5822.28\,m\,(5.822\,km)$

Explanation:

La distancia recorrida es la suma de las etapas de aceleración y velocidad constante (Travelled distance is the sum of the acceleration and constant speed stages). La aceleración es (The acceleration is):

$a = \frac{4.611\,\frac{m}{s}-0\,\frac{m}{s}}{360\,s}$

$a = 0.013\,\frac{m}{s^{2}}$

La distancia recorrida es (The travelled distance is):

$s = \frac{1}{2}\cdot \left(0.013\,\frac{m}{s^{2}} \right)\cdot (360\,s)^{2} + \left(4.611\,\frac{m}{s} \right)\cdot (1080\,s)$

$s = 5822.28\,m\,(5.822\,km)$

3 0

3 years ago