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erastova [34]
3 years ago
15

Before the start of a trip, air in a tire is at 320 kPa gage pressure and 27°C temperature. At the end of the trip the tire pres

sure is 360 kPa gage. Atmospheric pressure is 100 kPa. What is the temperature of air in the tire at the end of the trip in °C ? (Assume the air volume in the tire is unchanged, no air leaks out, and the air has ideal gas properties)…
Physics
1 answer:
tia_tia [17]3 years ago
7 0

Answer:

The temperature of air in the tire is 55.57 ºC

Explanation:

Please look at the solution in the attached Word file

Download docx
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What kind of energy does a skier have standing still at the top of a hill?
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The skier has potential because potential energy is enery that is stored or an object that is or does not move
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Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual
Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0
2 years ago
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in th
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Read 2 more answers
At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field
vesna_86 [32]

Answer:

9.7\times 10^{-5} T

Explanation:

Velocity =5\times 10^4i-2\times 10^4j

r=0.03i+0.05j

r=\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058

v=\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}

We know that

B=\frac{mv}{qr}

Where q=1.6\times 10^{-19} C

Mass of proton=1.67\times 10^{-27} kg

Using the formula

B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}

B=9.7\times 10^{-5} T

3 0
3 years ago
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