The bending of wave crests as the reach shallow water is:

A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure

Alex787 [66]

V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

   t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
   Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

   v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
   speed = |-2| m/s = 2 m/s

Moving right
   V > 0 => t^2 - 9t + 18 > 0
   (t - 6)(t - 3) > 0

   => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)



Going faster and slowing dowm

acceleration, a = v' = 2t - 9
   a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
   Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)

4 0

3 years ago

The heat flux that is applied to the left face of a plane wall is q" = 20 w/m2. the wall is of thickness l = 10 mm and of therma

inessss [21]

Of course steady state condition occurs in almost any system but time it will occurs varies among system. for this kind of system, conduction, steady state conduction occurs when the temperature change from one point to the point is already constant. steady state is not achieved immediately because the heat travels and material will not be heated at the same way at the starting point.

5 0

3 years ago

Read 2 more answers

The driver then tests the brakes on the car and safely comes to a complete stop with constant acceleration from 26.8 meters per

Temka [501]

Answer:

62.78

Explanation:

5 0

3 years ago

25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression

levacccp [35]

maximum speed of cheetah is

$v_1 = v_{max}$

speed of gazelle is given as

$v_2 = v_{g}$

Now the relative speed of Cheetah with respect to Gazelle

$v_{12} = v_1 - v_2$

$v_{12} = v_{max} - v_g$

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

$d = v_{12}* t$

so by rearranging the terms we can say

$t = \frac{d}{v_{12}}$

$t = \frac{d}{v_{max} - v_g}$

so above is the relation between all given variable

6 0

4 years ago

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

3 years ago