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3 years ago

Where is the 3rd quadrant

Physics

1 answer:

jeka943 years ago

4 0

The 3rd Quadrant is the lower left one on a graph,
where ' x ' and ' y ' are both negative.

Your picture is completely unrelated to this question.

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A 1150 kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 7.69 m before contacting the beam,

Natasha_Volkova [10]

Answer:

the average force 11226 N

Explanation:

Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.

Let's start looking for the stack speeds, it has a free fall, from rest (Vo=0)

Vf² = Vo² - 2gY

Vf² = 0 - 2 9.8 7.69 = 150.7

Vf = 12.3 m / s

This is the speed that the battery likes when it touches the beam. They also give us the distance it travels before stopping, let's calculate the time

Vf = Vo - g t

0 = Vo - g t

t = Vo / g

t = 12.3 / 9.8

t = 1.26 s

This is the time to stop

Now let's use the equation that relates the impulse to the amount of movement

I = Δp

F t = pf-po

The amount of final movement is zero because the system stops

F = - po / t

F = - mv / t

F = - 1150 12.3 / 1.26

F = -11226 N

This is the average force exerted by the stack on the vean

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3 years ago

Help me and you will meet your favorite singer at your door.

Goshia [24]

Sunlight, because it provides a source of energy is the answer because plants also provide a food source (please put as brainliest answer)

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3 years ago

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How did people studying lunar eclipses learn that Earth is round?

qwelly [4]

Answer:

Scientists have studied eclipses since ancient times. Aristotle observed that the Earth's shadow has a circular shape as it moves across the moon. He posited that this must mean the Earth was round. Another Greek astronomer named Aristarchus used a lunar eclipse to estimate the distance of the Moon and Sun from Earth

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3 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.

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

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3 years ago

Estimate the average rate of change in elevation from d to points i, ii, and iii, assuming the distance from d to i is 3500 m, t

Anit [1.1K]

Ummm i am not going to be able say i am high

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3 years ago