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2 years ago

15

The posted speed limit on the road heading from your house to school is45 mi/h, which is about 20 m/s. If you live 8 km (8,000 m

) from school, how long will it take you to get to school if there is no traffic? Convert your answer to minutes.

2 answers:

pav-90 [236]2 years ago

8 0

Answer:

6.67 min

Explanation:

Distance = speed × time

8000 m = 20 m/s × t

t = 400 s

Convert to minutes:

400 s × (1 min / 60 s) = 6.67 min

zaharov [31]2 years ago

6 0

Answer:

20.

Explanation:

not. tráfic. is. 20. minuts.

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What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22

SIZIF [17.4K]

Answer:

$5.45\times 10^{-4}$ W

Explanation:

$T_{r}$ = Temperature of the room = 22.0 °C = 22 + 273 = 295 K

$T_{s}$ = Temperature of the skin = 33.0 °C = 33 + 273 = 306 K

$A$ = Surface area = 1.50 m²

$\epsilon$ = emissivity = 0.97

$\sigma$ = Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴

Rate of heat transfer is given as

$R = \epsilon \sigma A (T_{s}^{2} - T_{r}^{2})$

$R = (0.97)(5.67\times 10^{-8}) (1.50) ((306)^{2} - (295)^{2})$

$R = 5.45\times 10^{-4}$ W

3 0

2 years ago

somewhere between the earth and the moon is a point where the gravitational attraction of the earth is canceled by the gravitati

mote1985 [20]

<span>It's pretty easy problem once you set it up.

Earth------------P--------------Moon

"P" is where the gravitational forces from both bodies are acting equally on a mass m

Let's define a few distances.
Rep = distance from center of earth to P
Rpm = distance from P to center of moon
Rem = distance from center of earth to center of moon

You are correct to use that equation. If the gravitational forces are equal then

GMearth*m/Rep² = Gm*Mmoon/Rpm²

Mearth/Mmoon = Rep² / Rpm²

Since Rep is what you're looking for we can't touch that. We can however rewrite Rpm to be

Rpm = Rem - Rep

Mearth / Mmoon = Rep² / (Rem - Rep)²

Since Mmoon = 1/81 * Mearth
81 = Rep² / (Rem - Rep)²

Everything is done now. The most complicated part now is the algebra, so bear with me as we solve for Rep. I may skip some obvious or too-long-to-type steps.

81*(Rem - Rep)² = Rep²
81*Rep² - 162*Rem*Rep + 81*Rem² = Rep²
80*Rep² - 162*Rem*Rep + 81*Rem² = 0

We use the quadratic formula to solve for Rep:
Rep = (81/80)*Rem ± (9/80)*Rem
Rep = (9/8)*Rem and (9/10)*Rem

Obviously, point P cannot be 9/8 of the way to the moon because it'll be beyond the moon. Therefore, the logical answer would be 9/10 the way to the moon or B.

Edit: The great thing about this idealized 2-body problem, James, is that it is disguised as a problem where you need to know a lot of values but in reality, a lot of them cancel out once you do the math. Funny thing is, I never saw this problem in physics during Freshman year. I saw it orbital mechanics in my junior year in Aerospace Engineering. </span> sylent_reality · 8 years ago

8 0

2 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

<u /> $F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

<u>Force on particle A due to particles B & C:</u>

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

<u>Force on particle C due to particles B & A:</u>

<u /> $F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$ <u />

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

<u>Force on particle B due to particles C & A:</u>

<u /> $F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$ <u />

<u /> $F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$ <u />

<u /> $F_b=3.49\times 10^{-5}\ N$ <u />

3 0

2 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

2 years ago

There is a 120 V circuit in a house that is a dedicated line for the dishwasher, meaning the dishwasher is the only resistor on

aliya0001 [1]

The info given in the question:

Voltage= 120V

Current=18A

Now we have to find the resistance. To find it use the following formula:

V=IR

Now making R to be the subject of the formula

R=V/I

R=120/18

The answer is 6.67 ohms

As dishwasher is the only resistor in the line the voltage drop is going to be 120V. The resistance values determines the hindrance that is present in the circuit that opposes the free flowing electrons

7 0

2 years ago

Read 2 more answers