Which of the following best defines spring constant ?

Do you think that biofuel is the only solution to the problem of reducing our

I am Lyosha [343]

Answer:

f biofuel production and use reduces our consumption of imported fossil fuels, we may become less vulnerable to the adverse impacts of supply

Explanation:

4 0

2 years ago

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I need help with part D

natka813 [3]

1). The little projectile is affected by friction all the way through the block.
Friction robs some kinetic energy.

2). The block is affected by friction as it scrapes along the top of the post.
Friction robs some kinetic energy.

3). The block is also affected by friction with the air (air resistance) as it
falls to the ground. Friction robs some kinetic energy.

8 0

3 years ago

A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point

Mandarinka [93]

Answer:

$F=9.09\times 10^{-9} N$

Explanation:

We are given that

Mass of spherical shell, $m_1$ =1900 kg

Mass= $m_2=1.80 kg$

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

$F=G\frac{m_1m_2}{r^2}$

$G=6.67\times 10^{-11} Nm^2/kg^2$

Using the formula

$F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}$

$F=9.09\times 10^{-9} N$

Hence,the gravitational force = $F=9.09\times 10^{-9} N$

6 0

2 years ago

Lemonade is best described as _______. A. an element B. a compound C. a mixture D. a pure substance

Minchanka [31]

C is the answer because it mix and water combining together

7 0

2 years ago

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Enrico Fermi (1901–1954) was a famous physicist who liked to pose what are now known as Fermi problems, in which several assumpt

Katarina [22]

Answer:

Explanation:

(a)

Since the earth is assumed to be a sphere.

Volume of atmosphere = volume of (earth +atm osphere) — volume of earth

$= \frac{4}{3}\pi(6400+ 50)^3 - \frac{4}{3}\pi (6400)
^3\\\\= \frac{4}{3}\pi(6192125000) km’^3\\= 2.6\times 10^{19} m^3$

Hence the volume of atmosphere is $2.6\times 10^{19} m^3$

(b)

Write the ideal gas equation as foll ows:

$PV = nRT\\\\n\frac{0.20atm\times 2.6\times10^{19} m^3}{0.08206L\, atm/mok\, K \times (15+273+15)K}\times \frac{1L}{10^{-3}m^3}\\\\= 2.20\times 10^{20} moles$

$no.\, of\, molecules = 2.20\times 10^{20} moles \times \frac{6.022\times10^{23}\,molecules}{1mole}= 13.3\times10^{43} molecules
$

Hence the required molecules is $13.3\times10^{43} molecules
$

(c)

Write the ideal gas equation as follows:

$PV =nRT
\\\\n=\frac{1.0 atm \times 0.5L
}{0.08206 L\, atm/mol\,K \times (37 +273.1 5)K} = 0.0196 moles$

$no.\, of\, molecules = 0.0196 moles \times\frac{6.022\times10^{23} molecules}
{Imole}= 1.2\times 10^{23} molecules$

Hence the required molecules in Caesar breath is $1.2\times 10^{23} molecules$

(d)

Volume fraction in Caesar last breath is as follows:

$Fraction,\, X =\frac{12\times 10 molecules}{13.3\times 10^{43} \,molecules}= 9.0\times 10\, molecule/air\, molecule}$

(e)

Since the volume capacity of the human body is 500 mL.

$Volume\, of\, Caesar\, nreath\, inhale\, is =\frac{ 12\times 10^{22}\, molecules}{breath}\times \frac{9.0\times10^{-23} molecule}{air\, molecule}\\\\= 1.08 molecule/breath$

5 0

2 years ago