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2 years ago

¿Por qué si cargas a uno de tus compañeros por cierto tiempo no estás realizando un trabajo mecánico?

Physics

1 answer:

VladimirAG [237]2 years ago

4 0

Answer:

I will answer this in English, we can translate it to:

Why if you charge a mate by an amount of time you are not doing work?

This happens because work is defined as the displacement done by a force:

W = d*F

where W is work, d is the distance, and F is the force.

This means that the amount of time that you are charging your mate does not affect the mechanical work, the only time that you are doing work is when you are lifting him.

You might be interested in

The thinnest layer of the earth is the:

Serhud [2]

Answer:

crust

Explanation:

it's the outermost layer of the planet.

7 0

2 years ago

Read 2 more answers

Consider a semi-spherical bowl machined out of wood (A drawing of a semi-spherical bowl can be found in UNIT- 4 Introduction, so

Mrrafil [7]

Answer:

The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)

Explanation:

The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.

a) Outer semi-sphere:

A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²

b) Inner semi-sphere:

A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²

c) Edge (Ring):

A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²

Therefore, the total surface area of the bowl is given by:

A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)

Changing units to m², as required in the problem, we get:

A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)

5 0

2 years ago

An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

NemiM [27]

1. Define Newtons second law of motion (this will help put things into perspective)

2.Get the mass of the object (in this case 75 kg)

3.The net force acting on the object...find it (in this case 500 N)

4.Change the equation to F=ma (500=75a)

5.Divide both sides by 75 and that is the acceleration.

7 0

2 years ago

Read 2 more answers

A certain frictionless simple pendulum having a length l and mass m swings with period t. If both l and m are doubled, what is t

iVinArrow [24]

If l and m both are doubled then the period becomes √2*T

what is a simple pendulum?

It is the one which can be considered to be a point mass suspended from a string or rod of negligible mass.

A pendulum is a weight suspended from a pivot so that it can swing freely.

Here,

A certain frictionless simple pendulum having a length l and mass m

mass of pendulum = m

length of the pendulum = l

The period of simple pendulum is:

$T = 2\pi \sqrt{\frac{l}{g} }$

Where k is the constant.

Now the length and mass are doubled,

m' = 2m

l' = 2l

$T' = 2\pi \sqrt{\frac{2l}{g} }$

$T' = \sqrt{2}* 2\pi \sqrt{\frac{l}{g} }$

$T' = \sqrt{2} * T$

Hence,

If l and m both are doubled then the period becomes √2*T

Learn more about Simple Harmonic Motion here:

<u>brainly.com/question/17315536</u>

#SPJ4

8 0

1 year ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

2 years ago