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kow [346]
3 years ago
7

A man is jogging at 3 m/s. If his mass is 85 kg, what is his kinetic energy?

Physics
1 answer:
Lina20 [59]3 years ago
8 0

Answer:

This is the complete answer and step by step explanation.

Given:

a = 3m/s

m = 85 kg

Unknown:

The Kinetic energy of the man jogging

Formula and solution:

KE = 1/2 mv²

KE = 1/2(85 kg)(3 m/s)²

KE =  (42.5 kg)(9 m²/s²)

Final Answer:

KE = 382.5 kg.m²/s² or simplify as 382.5 J is the kinetic energy of man jogging

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Lana71 [14]

Answer:

no

Explanation:

this is because its valency shell is full so it wont want any other electrons in its valence shell.

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Identify the solutions (homogeneous mixtures) in the list below. check all that apply. lead, an alloy of tin and lead ammonia, a
Goryan [66]

The solutions and homogeneous mixtures in the given list are:

A. Lead solder, an alloy of tin and lead.

D. Window cleaner, a mixture of ammonia and coloring dissolved in water.

E. Gasoline, a mixture of organic liquids with a fixed composition throughout.

<h3>What is a solution?</h3>

A solution can be defined as a special type of homogeneous mixture that comprises a solute and a solvent.

A homogeneous mixture can be defined as any solid, liquid, or gaseous mixture which has an identical (uniform) composition and properties throughout any given sample of the mixture.

In Science (Physics), all solutions are considered to be a homogeneous mixture because their constituents are uniformly (evenly) distributed.

In conclusion, the solutions and homogeneous mixtures in the given list are:

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  • Window cleaner, a mixture of ammonia and coloring dissolved in water.
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brainly.com/question/22070951

4 0
2 years ago
Read 2 more answers
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

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Answer:

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