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3 years ago

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1. what information does the atomic mass of an element provide?

2 answers:

avanturin [10]3 years ago

8 0

1. d
2. a
in an atom, the protons=electrons
3. a
atomic number= number of protons
4. c
protons are the "fingerprint" of an element
5. d
hope this helps!

Vilka [71]3 years ago

3 0

D, D, C, C, D.
There is no such designation as "Brainly pro", and no mechanism for awarding more points after an answer is posted. But thanks for the insanely generous 1 full point per answer anyway !

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On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

a)

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

b)

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

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3 years ago

How far will a freely falling object fall from rest in 5 seconds?

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<h2><em>how far will a freely falling object fall from rest in 5 seconds?</em></h2>

<em>If an object free falls from rest for 5 seconds, its speed will be <u>about 50 m/s.</u></em>

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A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of

Roman55 [17]

Answer:

$T=9.42Nm$

Explanation:

From the question we are told that:

Radius $r= 0.5 m$

Current $I= 3.0 A$

Normal vector $n=\frac{(2i - j +2k)}{3}$

Magnetic field $B= (2i-6j) T$

Generally the equation for Area is mathematically given by

$A=\pi r^2$

$A=3.1415 *0.5^2$

$A=0.7853 m^2$

Generally the equation for Torque is mathematically given by

$T=A(i'*B)$

Where

$i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}$

$X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]$

$X\ component\ of\ i'*B=12$

Therefore

$T=0.7853*12$

$T=9.42Nm$

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