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dezoksy
1 month ago
5

13. An aircraft heads North at 320 km/h rel:

Physics
1 answer:
Yuliya22 [3.3K]1 month ago
7 0

The aircraft's velocity relative to the ground is 240 km/h toward North

Explanation:

This problem can be addressed using vector addition. Essentially, the aircraft's velocity relative to the ground is the (vector) total of the aircraft's velocity through the air combined with the air's velocity relative to the ground.

Mathematically:

v' = v + v_a

where

v' denotes the aircraft's velocity relative to the ground

v represents the aircraft's velocity concerning the air

v_a indicates the air's velocity concerning the ground.

Considering north as the positive direction, we ascertain:

v = +320 km/h

v_a = -80 km/h (given the air is moving from North)

Thus, we calculate

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

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Kindly use English in your post so that I or someone else can assist you.
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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
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Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

\texttt{ }

Additional explanation

The Ideal Gas Law that should be remembered is:

\large {\boxed {PV = nRT} }

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

\texttt{ }

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

Next, we will determine the work performed on the gas:

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

Step B:

By utilizing the methodology mentioned earlier:

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

Next, we will ascertain the work completed on the gas:

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

Ultimately, we can calculate the total work done and heat supplied as follows:

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

Learn more

  • Minimum Coefficient of Static Friction:
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\texttt{ }

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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1 month ago
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