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2 years ago

What forces do a free body diagram analyze?

1 answer:

5 0

Friction, gravity, normal force, drag force, tension force and human force are the forces that occurs on free body.

<h3>What are the forces in a free body diagram?</h3>

There are large number of external forces acting on an object such as friction, gravity, normal force, drag force, tension force and human force due to pushing or pulling. These forces can cause the motion of the free body which is present at rest form or can stop a body which is moving. A free-body diagram is a useful means of describing and analyzing all the forces that act on a body to determine equilibrium.

So we can conclude that friction, gravity, normal force, drag force, tension force and human force are the forces that occurs on free body.

Learn more about force here: brainly.com/question/388851

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A spring attached to a mass is at rest in the initial

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Answer:

E=(1/2)kx^2

Explanation:

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3 years ago

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A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ

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Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

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3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

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3 years ago

Suppose you increase your walking speed from 4 m/s to 13 m/s in a period of 3 s. What is your acceleration?

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The acceleration formula goes like this: a= (vf-vi)/t so it would be (13-4)/3 Thus the answer is 3m/s^2

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4 years ago

A 2 L balloon filled with gas is warmed from 280 K to 700 K. What is the volume of the gas after it is heated?

irakobra [83]

Answer:

New volume, v2 = 0.8L

Explanation:

<u>Given the following data;</u>

Original Volume = 2L

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To find new volume V2, we would use Charles' law.

Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Charles is given by;

$VT = K$