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Gwar
2 months ago
8

A water-skier with weight Fg = mg moves to the right with acceleration a. A horizontal tension force T is exerted on the skier b

y the rope, and a horizontal drag forcc Fa is cxertcd by thc watcr on the ski. The water also exerts a vertical lift force L on the skier. Which of the following are correct relationships between the forces exerted on the skier-ski system? Select two answers. (A) T-Fd = ma (B) L-Fg=ma (C) L- =0 (D) T- Fd 0
Physics
1 answer:
Maru [3.3K]2 months ago
3 0

Answer:

The accurate equations are T-fg=ma and L-fg=0.

Options (A) and (C) are valid.

Explanation:

Given that:

Weight Fg = mg

Acceleration = a

Tension = T

Drag force = Fa

Vertical force = L

We are required to determine the correct relationships

Using the equilibrium equations

Horizontally,

The acceleration is a

T-Fd=ma...(I)

Vertically,

No acceleration occurs

w=L

mg-L=0

Substituting the value of mg

L-fg=0....(II)

Thus, the correct relationships are T-fg=ma and L-fg=0.

Options (A) and (C) are correct.

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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)

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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

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\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}

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

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

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

\texttt{ }

Learn more

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\texttt{ }

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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