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aleksandr82 [10.1K]
3 years ago
6

the air in a tire is initially at 380 kpa, 20 C, when the tire volume is 0.120 m^3. as the tire is warmed by the sun, the pressu

re becomes 450 kpa and the tire volume increases by 5%. determine the final temperature and the mass of air in the tire.
Physics
1 answer:
castortr0y [4]3 years ago
7 0

Answer:

final temperature is 364.32 K

mass of air is 0.5423 kg

Explanation:

given data

pressure p1 = 380 kPa

volume v1 =  0.120 m³

temperature = 20°C = 20 + 273 = 293 K

pressure p2 = 3450 kPa

volume v2 =  5% increase

to find out

final temperature and the mass

solution

we consider here ideal gas

so equation is

p1v1 / t1 = p2v2/t2

put here all value and find t2

and v2 is = 0.120 ( 1 + 0.05) = 0.126 m³

so

t2 = p2v2/ p1v1 × t1

t2 = (450 × 0.126) /  (380 × 0.120)  ×  293

t2 = 364.32 K

so final temperature is 364.32 K

and

mass = p2v2 / Rt2

here R gas constant is 0.287 kJ/kg.K

so

mass = (450 × 0.126) / (0.287 × 364.32 )

mass = 0.5423

so mass of air is 0.5423 kg

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Brilliant_brown [7]

Answer:

-96.465m joule

Explanation:

Let m = mass of the car and v1 = initial velocity and v2 = final velocity

Given.

Initial velocity = 100 km/h

final velocity = 50 km/h

What is work done in the car to slow it from 100km/h to 50km/h?

v1=100km/h=27.78m/s

v2=50km/h=13.89m/s

The work done in the car to slow it from v1 to v2.

w=Δk

w=k2-k1

w= \frac{1}{2}m(v2)^{2}- \frac{1}{2}m(v1)^{2}

w=\frac{1}{2} m(v2-v1)^{2}

w=\frac{1}{2}\times m(13.89 -27.78)^{2}

w=\frac{1}{2}\times m(-13.89)^{2}

w=\frac{1}{2}\times m\times (-192.93)

w=-96.465m joule.

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7 0
3 years ago
Group one on the periodic table shares which characteristics.?
Dmitry_Shevchenko [17]
Group one on the periodic table, also known as the alkali metals, all:
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4 0
3 years ago
Which of the following is a way that some animals bodies adapt to living in extremely cold climates A. Clusters B. Hibernation C
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3 0
4 years ago
ichrome wire of cross-sectional radius 0.791 mm is to be used in winding a heating coil. If the coil must carry a current of 9.2
77julia77 [94]

Answer:

length is 23.228091 m

Explanation:

Given data

radius = 0.791 mm = 0.791 × 10^{-3} m

current =  9.25 A

voltage =  1.20 × 10² V

to find out

resistance and  length of wire

solution

we know the resistance formula that is

resistance = voltage / current

resistance = 1.20 × 10²  /9.25

resistance = 12.97

so resistance is 12.97 ohm

and length of nichrome wire formula is

length = resistance × area /  specific resistance of wire  

so specific resistance of wire  we know = 1.1 × 10^{-6}

and area = \pi × r² = \pi × (0.0791× 10^{-3})²

area = 1.97 × 10^{-6}

length =   12.97  × 1.97 × 10^{-6}  /  1.1 × 10^{-6}

so length is 23.228091 m

6 0
3 years ago
13. Which best summarizes the change in number and size of microcontinents during the Proterozoic?
Nutka1998 [239]

Answer:

Option C

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Explanation:

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7 0
3 years ago
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