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natulia [17]
3 years ago
12

a lemon with mass 0.3 kg falls out of a tree from a height of 1.8 m. How much mechanical energy does the lemon have just before

it hits the ground ?
Physics
2 answers:
Nadusha1986 [10]3 years ago
6 0
5.30 joules of mechanical energy. Hope I helped!
cestrela7 [59]3 years ago
6 0

Answer:

5.292 Joules

Explanation:

Hello

applying the law of the conservation of energy, and assuming that no energy is lost due to friction with the air, we can affirm that the energy that has just before the impact is the same energy that it has just before being released, that energy is due to height, mass and gravity and is defined by:

E_{g}=m*g*h\\\\now\\E_{p} =E_{g}=m*g*h\\E_{p}=m*g*h\\

Let

m=0.3kg\\g=9.8\frac{m}{s^{2} }\\ h=1.8\ m

pute the values into the equation

E_{p}=m*g*h\\E_{p}=0.3 kg*9.8\frac{m}{s^{2} }*1.8\m\\E_{p}=5.292\ Joules\\

Have a nice day.

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Answer: Speeding Up

Explanation:

Force is proportional to the acceleration of an object. The greater the force, the more acceleration will be produced.

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Coulomb's law states that the force F of attraction between two oppositely charged particles varies jointly as the magnitude of
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Explanation:

Let the charges be q1 and q2 and the distance between the charges be 'd'

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F1=kq1q2/d²...(1)

Where k is the electrostatic constant.

If q1 and q2 is doubled and the distance halved, we will have;

F2 = k(2q1)(2q2)/(d/2)²

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3 years ago
Instantaneous speed is measured
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Answer:

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Explanation:

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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 added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

\texttt{ }

<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 !

\texttt{ }

<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>

<h3>Step A:</h3>

<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{ }

<h3>Step B:</h3>

<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

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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Explanation:

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