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3 years ago

What is the purpose of a free body diagram

2 answers:

4 0

<span>A free body diagram is a representation of how the forces that are acting on a point or particle interact. You place your point at the origin and then draw your forces.

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Inga [223]3 years ago

4 0

Answer:

The purpose of a free body diagram is to make a visual representation of all the forces acting on an object.

In a free-body diagram, the object is represented at the centre; every force acting on the object is represented as an arrow, starting from the centre of the object, and with a length proportional to the magnitude of the force and direction equal to the direction of the corresponding force.

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Which sector is not part of the circular flow model of the economy ?

morpeh [17]

Answer:

a. government sector

Explanation:

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3 years ago

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The length of the track itself is 500m. The vehicles (each of mass 15kg) are dragged

gogolik [260]

Answer:

W = 100000 J = 100 KJ

Explanation:

Here we will use the most basic and general formula of work, which is as follows:

$W = Fd$

where,

W = Work Done = ?

F = Force Required = 200 N

d = Length of Track = 500 m

Therefore,

$W = (200\ N)(500\ m)\\$

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2 years ago

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3 years ago

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) Un círculo de 120 cm de radio gira a 600 rpm. Calcula: a) su velocidad angular

DIA [1.3K]

Responder:

20πrads ^ -1; 24πrads ^ -1; 0,1 seg; 10 Hz

Explicación:

Dado lo siguiente:

Radio (r) del círculo = 120 cm

600 revoluciones por minuto en radianes por segundo

(600 / min) * (2π rad / 1 rev) * (1min / 60seg)

(1200πrad / 60sec) = 20π rad ^ -1

Velocidad angular (w) = 20πrads ^ -1

Velocidad lineal = radio (r) * velocidad angular (w)

Velocidad lineal = (120/100) * 20πrad

Velocidad lineal = 1.2 * 20πrads ^ -1 = 24πrads ^ -1

C.) Período (T):

T = 2π / w = 2π / 20π = 0.1 seg

D.) Frecuencia (f):

f = 1 / T = 1 / 0.1

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2 years ago

In a tank full of water, the pressure on a surface 2 meters below the water level is 1.5 kPA. What's the pressure on a surface 6

Lina20 [59]

Answer:

P = 40.7kPa

Explanation:

To find the pressure on a surface 6 meter below you use the following formula, which takes into account the heights in which pressures are measured and also the density of the fluid and the gravitational acceleration:

$P_2-P_1=-\rho g(y_2-y_1)$ (1)

P2: pressure for a height of -6 m = ?

P1: pressure for a height of -2 m = 1.5kPa = 1500 Pa

ρ: density of water = 1000kg/m^3

g: gravitational acceleration = 9.8 ms^2

y2: -6m

y1: -2m

(the height is measure from the water level, because of that, the heights are negative)

You solve the equation (1) for P1:

$P_1=P_2-\rho g(y_2-y_1)$ (2)

Next, you replace the values of all variables in equation (2):

$P_2=1500Pa-(1000kg/m^3)(9.8m/s^2)(-6-(-2))m=40700Pa\\\\P_2=40.7kPa$

hence, the pressure on a surface 6 m below the water level is 40.7kPa

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2 years ago