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goblinko [34]
4 years ago
7

What is the difference between: first moment of area and second moment of area?

Physics
1 answer:
zhannawk [14.2K]4 years ago
6 0
To determine the centroid of the object first moment of area is used.

To predict the resistance of a shape to bending and deflection which are directly proportional, second moment of area is used.

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Consider three drinking glasses. All three have the same base area, and all three are filled to the same depth with water. Glass
natulia [17]
The answer to this question is D
7 0
3 years ago
If a person weighs 600N on earth, how much does he weigh on the moon
Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0
4 years ago
A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle
emmasim [6.3K]

<u>Given data:</u>

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping <em>μs = tan Φ, Φ = Θ </em>

<em>Pmin = W sin( α +Θ  )</em>

<em>          = W[ sin α.Cos Θ + cos α.sin Θ]</em>

<em>           = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]</em>

<em>           = 460.9 N</em>

<em>Minimum push required to move the box up the ramp is 460.9 N</em>


7 0
4 years ago
Which of the following statements about features on the sun is true
pashok25 [27]
D is the answer because of the solar phonominals
7 0
3 years ago
A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the
Julli [10]

Answer:

<h2>a) Q = 0.759µC</h2><h2>b) E = 39.5µJ</h2>

Explanation:

a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV

C = capacitance of the capacitor (in Farads )

V = voltage (in volts) = 100V

C = ∈A/d

∈ = permittivity of free space = 8.85 × 10^-12 F/m

A = cross sectional area = 600 cm²

d= distance between the plates = 0.7cm

C = 8.85 × 10^-12 * 600/0.7

C = 7.59*10^-9Farads

Q = 7.59*10^-9 * 100

Q = 7.59*10^-7Coulombs

Q = 0.759*10^-6C

Q = 0.759µC

b) Energy stored in a capacitor is expressed as E = 1/2CV²

E = 1/2 * 7.59*10^-9 * 100²

E = 0.0000395Joules

E = 39.5*10^-6Joules

E = 39.5µJ

7 0
3 years ago
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