Write with all the steps and formulas and drawing if needed.

Physics

1 answer:

lianna [129]3 years ago

7 0

<u>The answer is not contained detail explanation, just a solution and the required values. </u>

All the details are in the pictures, the answers are marked with orange colour.

Note,

in the task no 20.:

$m_A - the \ mass \ of \ A; \ m_B-the \ mass \ of \ B \ balls.\\V_A \ and \ V_B-the \ velocities \ of \ the \ A&B \ balls \ before \ collision.\\V'_A \ and \ V'_B-the \ velocities \ of \ the \ A&B \ balls \ after \ collision.$

V - the velocity of the pair of the balls after collision.

in the task no 21:

m₁ - the mass of the copper ball; m₂ - the mass of the copper calorimeter; m₃ - the mass of the water; t₀ - the initial temperature of water in the copper calorimeter; θ - the final temperature in the calorimeter after the copper ball is transferred into a copper calorimeter; t₁ - the required initial temperature of the copper ball before it is transferred into the calorimeter.

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Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =

Yanka [14]

The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>

4 0

3 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations: