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tiny-mole [99]
3 years ago
12

What is centrifugal force with respect to unbalance? What is the formula used to determine centrifugal force?

Engineering
1 answer:
3241004551 [841]3 years ago
4 0

Answer:

F = \frac{m * v^2}{r}

Explanation:

Once the centrifugal forces are equivalent as an object rotates, it is called "in equilibrium." If the point of impact of the moving object does not align with the center of the rotation, unequal centrifugal forces are produced and the rotating component is "out of equilibrium."  

Two conditions for balanced condition are: static and coupled. Static equilibrium relates to a one plane mass in which the unit is balanced by  weight inverse the unbalanced mass

formula used to calculate centrifugal mass is F = \frac{m * v^2}{r}

where, m is mass, v is speed of body. r is radius

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All of the dimensions on an aircraft drawing are_________<br> to the bottom of the drawing.
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All of the dimensions on an aircraft drawing are _________ to the bottom of the drawing


Answer: parallel
7 0
2 years ago
What is the least count of screw gauge?<br> (a) 0.01 cm<br> (b) 0.001 cm<br> (c) 0.1 cm<br> (d) 1 mm
Nonamiya [84]
Its 0.001

0.01 x100 = 1mm
0.001x100=0.1mm
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1m
3 0
2 years ago
I have to find the critical points of this function of two variables <img src="https://tex.z-dn.net/?f=%5C%5Cf%28x%2Cy%29%3Dx%5E
liraira [26]

Answer:

no i dont think there is

Explanation:

because theres not

4 0
3 years ago
The drag coefficient of a car at the design conditions of 1 atm, 25°C, and 90 km/h is to be determined experimentally in a large
SIZIF [17.4K]

Answer: 0.288

Explanation:

Given

Pressure of the car, P = 1 atm

Temperature of the car, T = 25° C

Speed of the car, v = 90 km/h = 90*1000/3600 = 25 m/s

Height of the car, h = 1.25 m

Width of the car, b = 1.65 m

Force acting on the far, F = 220 N

Drag coefficient, C(d) = ?

Using our table A-9, we can trace that the density of air ρ, at the given temperature and pressure of 25 °C and 1 atm, is 1.184 kg/m³

Area = h *b

Area = 1.25 * 1.65

Area = 2.0625 m²

Now we solve for the drag coefficient using the formula

C(d) = F / (1/2 * ρ * A * v²)

C(d) = 220 / (0.5 * 1.184 * 2.0625 * 25²)

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C(d) = 0.288

Therefore, the drag coefficient is 0.288

3 0
2 years ago
Technician A says that forma hardened steel may have different strenght áreas. Technician B says that aluminum collapses in a pr
stich3 [128]

Technician B is correct because the way aluminum collapses can be predicted.

Hardened steel and aluminum are two metals used for different purposes including:

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  • Appliances.
  • Small utensils.
  • Airplanes.
  • Vehicles.

These two materials have slightly different features in terms of resistance, flexibility, etc.

In the case of hardened steel, this is considered to be malleable but strong. This means it is possible to change its shape under some conditions but it can resist great forces and pressure. Moreover, if the hardening process is carried out properly all the areas should be equally strong.

On the other hand, aluminum is recognized due to its durability and for being lighter than other materials. Despite this, aluminum is more flexible than steel and collapses under weaker forces. This has been widely studied because aluminum collapse shows a predictable pattern.

Based on this, only technician B is correct.

Learn more in: brainly.com/question/24043240

5 0
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