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kolbaska11 [484]

3 years ago

12

An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40

'10", and 35 ∘ 39'55"A) Calculate the standard deviation.B)Calculate the standard deviation of the mean.

1 answer:

Helen [10]3 years ago

6 0

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

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The driveshaft of an automobile is being designed to transmit 238 hp at 3710 rpm. Determine the minimum diameter d required for

AURORKA [14]

Explanation:

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3 years ago

Pipe (2) is supported by a pin at bracket C and by tie rod (1). The structure supports a load P at pin B. Tie rod (1) has a diam

Galina-37 [17]

Answer:

P_max = 25204 N

Explanation:

Given:

- Rod 1 : Diameter D = 12 mm , stress_1 = 110 MPa

- Rod 2: OD = 48 mm , thickness t = 5 mm , stress_2 = 65 MPa

- x_1 = 3.5 mm ; x_2 = 2.1 m ; y_1 = 3.7 m

Find:

- Maximum Force P_max that this structure can support.

Solution:

- We will investigate the maximum load that each Rod can bear by computing the normal stress due to applied force and the geometry of the structure.

- The two components of force P normal to rods are:

Rod 1 : P*cos(Q)

Rod 2: - P*sin(Q)

where Q: angle subtended between x_1 and Rod 1 @ A. Hence,

Q = arctan ( y_1 / x_1)

Q = arctan (3.7 / 2.1 ) = 60.422 degrees.

- The normal stress in each Rod due to normal force P are:

Rod 1 : stress_1 = P*cos(Q) / A_1

Rod 2: stress_2 = - P*sin(Q) / A_2

- The cross sectional Area of both rods are A_1 and A_2:

A_1 = pi*D^2 / 4

A_2 = pi*(OD^2 - ID^2) / 4

- The maximum force for the given allowable stresses are:

Rod 1: P_max = stress_1 * A_1 / cos(Q)

P_max = (110*10^6)*pi*0.012^2 / 4*cos(60.422)

P_max = 25203.61848 N

Rod 2: P_max = stress_2 * A_2 / sin(Q)

P_max = (65*10^6)*pi*(0.048^2 - 0.038^2) / 4*sin(60.422)

P_max = 50483.4 N

- The maximum force that the structure can with-stand is governed by the member of the structure that fails first. In our case Rod 1 with P_max = 25204 N.

8 0

2 years ago

Where are the ar manufacturers not fitting the engine in the high end sport cars

fomenos

Answer:

it depends on the but i would recommend check in the front next to the turbo intake.

8 0

2 years ago

(a) The lower yield point for an iron that has an average grain diameter of 1 x 10-2 mm is 230 MPa. At a grain diameter of 6 x 1

olya-2409 [2.1K]

Answer:

The answer is " $4.35 \times 10^{-3}\ mm$ and 157.5 MPa".

Explanation:

In point A:

The strength of its products with both the grain dimension is linked to this problem. This formula also for grain diameter of 310 MPA is represented as its low yield point

$y = yo + \frac{k}{\sqrt{x}}$

Here y is MPa is low yield point, x is mm grain size, and k becomes proportionality constant.

Replacing the equation for each condition:

$y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k$

People can get yo = 275 MPa with both equations and k= 15.5 Mpa $mm^{\frac{1}{2}}$ .

To substitute the answer,

$310 = 275 + \frac{(15.5)}{\sqrt{x}}\\\\x = 0.00435 \ mm = 4.35 \times 10^{-3}\ mm$

In point b:

The equation is $\sigma y = \sigma 0 + k y d^{\frac{1}{2}}$

equation is:

$75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\$

by putting the above value in the formula we get the $\sigma y$ value that is= 157.5 MPa

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2 years ago

What is an electrical output device

tester [92]

Answer:

heyoo!

a printer, camera, computerr

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Explanation:

7 0

2 years ago

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