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Pavlova-9 [17]
3 years ago
10

Um ônibus percorre a distância de 480 km, entre Santos e Curitiba, com velocidade escalar média de 80 km/h. De Curitiba a Floria

nópolis, distantes 300 km, o ônibus desenvolve a velocidade escalar média de 75 km/h. Qual a velocidade escalar média do ônibus entre Santos e Florianópolis?
Physics
2 answers:
Mariulka [41]3 years ago
7 0

Answer:

10 h

Explanation:

velocidade é a taxa de variação da distância no tempo. é a razão entre a distância e o tempo

de Santos e Curitiba:

distância (d) de 480 km, velocidade (s) de 80 km/h

s=\frac{d}{t}\\ t=\frac{d}{s} =\frac{480}{80}=6h

de Curitiba e Florianópolis:

distância (d) de 300 km, velocidade (s) de 75 km/h

s=\frac{d}{t}\\ t=\frac{d}{s} =\frac{300}{75}=4h

tempo médio de ônibus entre Santos e Florianópolis = 6h + 4h = 10h

WITCHER [35]3 years ago
6 0

Answer:

78 km/h

Explanation:

To solve this problem, we need to find the total distance and time travelled, then we can divide them to find the average speed.

From Santos and Curitiba, the time travelled is:

480 = 80 * t1

t1 = 480 / 80 = 6 hours

From Curitiba to Florianopolis, the time travelled is:

300 = 75 * t2

t2 = 300 / 75 = 4 hours

So, the total time travelled is t1 + t2 = 10 hours

And the total distance travelled is 480 + 300 = 780 km

Then, we can calculate the average speed:

speed = 780 / 10 = 78 km/h

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7 0
3 years ago
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0
3 years ago
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