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3 years ago

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A driver in a car traveling at a speed of 26.8 m/s sees a deer 100 m away on the road. Calculate the minimum constant accelerati

on that is necessary of the car to stop without hitting the deer (assuming the deer does not move).
-4.20 m/s2

-1.75 m/s2

-2.89 m/s2

-3.59 m/s2

1 answer:

ZanzabumX [31]3 years ago

6 0

-3.5912m/s^2Use vf^2 = vi^2 + 2ax in your calculations, knowing vf, vi, and x.

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Use absolute dating in a sentence

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Answer:

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A jetliner can fly 4.9 hours on a full load of fuel. Without any wind it flies at a speed of 1.88 x 102 m/s. The plane is to mak

PSYCHO15rus [73]

Answer:

1371.26 Km

Explanation:

First of all, we need to find the velocity of the plane relative to the ground since the air has a velocity of 78.2 m/s due east and without any wind, it flies at a velocity of 188 m/s.

Thus, during the west trip, the velocity will be;

Vw = Vp - Va

Vp is velocity of plane while Va is velocity of air

and since distance/time = velocity ;

Time = velocity/distance and thus;

Time during this west period ;Tw = X/(Vp - Va)

Now during the east trip,

Ve = Vp + Va

And Te = X/(Vp + Va)

From the question, the plane can fly 4.9 hours on a full load of fuel. Let's convert this to seconds because velocity is in m/s

Thus, 4.9 hours = 4.9 x 60 x 60 = 17640 seconds

So, this time will be equal to the sum of that in the west and east directions.

Thus; T = Tw + Te

From above we know Tw and Te.

Let's substitute them into the equation;

T = [X/(Vp - Va)] + [X/(Vp + Va)]

T = X[(Vp + Va + Vp - Va)/((Vp)² — (Va)²)

T = X[(2Vp)/((Vp)² — (Va)²)

Making X the subject to obtain;

X = [T((Vp)² — (Va)²)]/(2Vp)

X = [17640((188)² — (78.2)²)]/(2 x 188)

X = 515595326.4/376 = 1371264.17m = 1371.26 Km

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3 years ago

You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it

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Answer:

30.25 m/s

46.68 m

Explanation:

Work Energy theorem states that

W = ½mv2² - ½mv1²

W = ½m(v2² - v1²)

Net work done by the force = -mgd

Net work done = -m * 9.8 * 14.8

Net work done = -145m

Using the work energy theorem

-145m = ½m(v2² - v1²)

-145m = ½ * m(25² - v1²)

-290m = 625m - v1²m

v1² = 625 + 290

v1² = 915

v1 = √915 = 30.25 m/s

B

-mgd = ½m(v2² - v1²), where v2 = 0, so

-mgd = ½mv1²

Making d the subject of the formula, we have

d = -½mv1²/mg

d = v1²/2g

d = 915/ 2 * 9.8

d = 915 / 19.6

d = 46.68 m

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Investigate the chemical potential μ upon a change in volume. We will use the fact that

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Answer:

$\mu_{Vf}-\mu_{Vi}=-3*10^{-21} J/mol$

Explanation:

Let's rewrite F in terms of N.

$F=32NkT+C-T(Nkln(V-bN)+C)$

$F=32NkT+C-TNk\ln{(V-bN)}-TC$

C is a constant value.

We know that the chemical potential μ is the partial derivative of F with respect to N.

$\frac{\partial F}{\partial N}=32kT-Tk(ln(V-bN)+\frac{N}{V−bN}(-b))$

$\frac{\partial F}{\partial N}=kT(32-(ln(V-bN)+\frac{N}{V−bN}(-b)))$

$\frac{\partial F}{\partial N}=kT(32-ln(V-bN)+\frac{Nb}{V−bN})$

We can find now the chemical potential μ(Vf).

$\mu_{Vf}=kT(32-ln(Vf-bN)+\frac{Nb}{Vf−bN})$

k is the Boltzmann constant $1.38*10^{-23} m^{2}kgs^{-2}K^{-1}$
N=20 moles
T is temperature 300 K
Vi initial volume 0.01 m3
Vf final volume 0.02 m3

$\mu_{Vf}=1.38*10^{-23}*300*(32-ln(0.02-(9*10^{-29}*20*6.022*10^{23}))+\frac{20*6.022*10^{23}*9*10^{-29}}{0.02-(9*10^{-29}*20*6.022*10^{23}}))$

$\mu_{Vf}= 1.49*10^{-19} J/mol$

We can do the same to find μ(Vi).

$\mu_{Vi}=1.38*10^{-23}*300*(32-ln(0.01-(9*10^{-29}*20*6.022*10^{23}))+\frac{20*6.022*10^{23}*9*10^{-29}}{0.01-(9*10^{-29}*20*6.022*10^{23}}))$

$\mu_{Vi}=1.52*10^{-19} J/mol$

Therefore µ(Vf) - µ(Vi) will be:

$\mu_{Vf}-\mu_{Vi}=-3*10^{-21} J/mol$

I hope it helps you!

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4 years ago