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3 years ago

How does the law of the conservation of mass relate to "You can't get something for nothing

1 answer:

8 0

You can't get something from nothing is practically stating the law. The law means states that energy can't be created or destroyed, in other words, it's constant. So in saying so, you truly can't get something from nothing

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The alternating current which crosses an apparatus of 600 W has a maximum value of 2.5 A. What is efficient voltage between its

azamat

Answer: Option (b) is correct.

Explanation:

Since we know that,

P = VI

where;

P = power

V= Voltage

I = Current

Since it's given that,

P = 600W

I = 2.5 A

equating these values in the above equation, we get;

8 0

2 years ago

I need help for a review sheet

siniylev [52]

Tin is Sn, atomic number 47 is Silver, Mass of sodium is 22.9 u

3 0

3 years ago

A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s

Ede4ka [16]

When acceleration is constant, the average velocity is given by

$\bar v=\dfrac{v+v_0}2$

where $v$ and $v_0$ are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

$\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}$

where $x,x_0$ are the final/initial displacements, and $t,t_0$ are the final/initial times, respectively.

Take the car's starting position to be at $t_0=0\,\mathrm s$ . Then

$\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t$

So we have

$x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m$

You also could have first found the acceleration using the equation

$v=v_0+at$

then solve for $x$ via

$x=x_0+v_0t+\dfrac12at^2$

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of $a$ anyway.

7 0

2 years ago

A freezer has a coefficient of performance of 4.1. You place 0.45 kg of water at 16°C in the freezer, which maintains its temper

velikii [3]

Ohh this is quite a lot umm i’ll be back hold on

5 0

2 years ago

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu

Setler79 [48]

<h2>Given :</h2>

total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

$\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

$\dfrac{10 {}^{19} }{1.6}$

$\dfrac{10\times 10 {}^{19} }{16}$

$\dfrac{100 \times 10 {}^{18} }{16}$

$\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

$0.009 \times 6.24 \times {10}^{18}$

$0.05616 \times 10 {}^{18}$

$\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

$\dfrac{5.616 \times {10}^{16} }{3.6}$

$\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

$10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

$\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

_____________________________

$\mathrm{ \#TeeNForeveR}$

6 0

2 years ago