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Nuetrik [128]
2 years ago
14

If the volume of the cylinder is to be calculated, what would be the total standard deviation of the volume?

Physics
1 answer:
Ahat [919]2 years ago
8 0

first calculate the partial derivatives of the two fromulas for each measured variable. Then you calculate the sum of the products of the errors (Dr, DR, and dh) with the squared corresponding partial derivative.and or the deviation

Example for the length of the mantle:

dm/dR = (R-r)/root(w)

dm/dr = -(R-r)/root(w)

dm/dh = h/root(w)

where w = (R-r)²+h². The squared derivatives are

(dm/dR)² = (R-r)²/w

When it comes to statistics and probability theory, standard deviation is used. It demonstrates the accuracy of your data and is used to measure both variability and diversity.

Standard deviation is calculated by taking the square root of the variance. In contrast to a high standard deviation, which indicates that the entered data points are most likely farther from the mean, a low standard deviation indicates that the entered data points are most likely closer to the mean

to learn more about deviation:

brainly.com/question/16555520

#SPJ4

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A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
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(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
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Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

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