A 1430 kg car speeds up from 7.5 m/s to 11 m/s in 9.3 s. Ignoring friction, how much power did that require?

Varvara68 [4.7K]

Answer:

D IS MY ANSWER N2 TO UREA HOPE ITS HELFUL

6 0

3 years ago

José and Laurel measured the length of a stick's shadow during the day. Without knowing the length of the stick, which of their

skelet666 [1.2K]

Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

5 0

3 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

When a puddle dries up what are the particles really doing

Alexandra [31]

The particles are either being absorbed or evaporating

7 0

3 years ago

A nonconducting sphere of radius 10 cm is charged uniformly with a density of 100 nC/m3. What is the magnitude of the potential

Flura [38]

Answer:

The answer is 3.0 V

Explanation:

the formula of the potential due to a non-conductive sphere with a uniform charge at a point at a distance x from the center of the sphere is equal to:

V1 = (2*π*k*ρ*(3*r^2 - x^2))/3

where

ρ = volume charge density = 100 nC/m^3

x = distance from center = 0

r = radius = 10 cm = 0.1 m

Thus:

V1 = (2*π*k*ρ*3*r^2)/3

for x = 4 cm = 0.04 m, we have:

V4 = (2*π*k*ρ*(3*r^2 - (0.04^2))/3 = (2*π*k*ρ*(3*r^2 - 0.0016))/3

The difference between both equations:

V1 - V4 = (2*π*k*ρ*3*r^2)/3 - (2*π*k*ρ*(3*r^2 - 0.0016))/3

V1 - V4 = (2*π*k*ρ*0.0016)/3 = (2*π*9x10^9*100x10^-9*0.0016)/3 = 3.02 V = 3 V

5 0

3 years ago