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2 years ago

A fox runs at a speed of 16 m/s and then stops to eat a rabbit. If this all took 120

Physics

1 answer:

GalinKa [24]2 years ago

7 0

Answer:

a = 52s²

Explanation:

<u>How to find acceleration</u>

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2). Acceleration is also a vector quantity, so it includes both magnitude and direction.

<u>Solve</u>

We know initial velocity (u = 16), velocity (v = 120) and acceleration (a = ?)

We first need to solve the velocity equation for time (t):

v = u + at

v - u = at

(v - u)/a = t

Plugging in the known values we get,

t = (v - u)/a

t = (16 m/s - 120 m/s) -2/s2

t = -104 m/s / -2 m/s2

t = 52 s

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Burka [1]

Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

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<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

$\large {\boxed {PV = nRT} }$

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

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<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<em>Ideal gas is allowed to expand isothermally:</em>

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

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

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_A = -P_2(V_2 - V_1)$

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

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

$\texttt{ }$

<em>Using the same method as above:</em>

$P_2V_2 = P_3V_3$

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

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

<em>Next we will calculate the work done on the gas:</em>

$W_B = -P_3(V_3 - V_2)$

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

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

$\texttt{ }$

<em>Finally we could calculate the total work done and heat added as follows:</em>

$W = W_A + W_B$

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

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

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

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

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

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

$\texttt{ }$

<h3>Learn more</h3>

Minimum Coefficient of Static Friction : brainly.com/question/5884009
The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 years ago

How r u?<br> Are u good?

Maslowich

Answer:

i'm good, hru hope ur staying safe

Explanation:

6 0

3 years ago

Read 2 more answers

Amazon rectangular bar of low carbon steel A-36 is exposed to an axial strees of 150 MPa. What is the original length of the bar

kolbaska11 [484]

Answer:

1.8m

Explanation:

Let the Elastics of the steel ASTM-36 $E = 200000 MPa$

The strain of the bar when subjected to 150 MPa is

$\epsilon = \frac{\sigma}{E} = \frac{150}{200000} = 0.00075$

Therefore, if the bar elongates by 1.35 mm, then the original length L would be:

$\epsilon = \frac{\Delta L}{L}$

$L = \frac{\Delta L}{\epsilon} = \frac{1.35}{0.00075} = 1800 mm$ or 1.8m

5 0

3 years ago

The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,

Nikolay [14]

Answer:

Index of expansion: 4.93

Δu = -340.8 kJ/kg

q = 232.2 kJ/kg

Explanation:

The index of expansion is the relationship of pressures:

pi/pf

The ideal gas equation:

p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

500 C = 773 K

20 C = 293 K

p2 = 35*0.1*773/(293*1.3) = 7.1 bar

The index of expansion then is 35/7.1 = 4.93

The variation of specific internal energy is:

Δu = Cv * Δt

Δu = 0.71 * (20 - 500) = -340.8 kJ/kg

The first law of thermodynamics

q = l + Δu

The work will be the expansion work

l = p2*v2 - p1*v1

35 bar = 3500000 Pa

7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

7 0

3 years ago

A 20 g sparrow flying toward a bird feeder mistakes the pane of glass in a window for an opening and slams into it with a force

Rama09 [41]

I found this using the app Socratic. When I took physics in high school it helped me so much.

5 0

3 years ago