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2 years ago

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Need help solving math problem using integration

1 answer:

notka56 [123]2 years ago

8 0

Ummm did you try to add or subtract and multiply or divide that can get your answer

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What’s the answer to this I don’t understand

kari74 [83]

Answer:

Electrons are the negatively charged particles of atom. Electrons are extremely small compared to all of the other parts of the atom. So, the negative green particles of this atom are the electrons.

Explanation:

Hope it helps:)

3 0

2 years ago

CO is the abbreviation for

garri49 [273]

Answer:

CO is usually the abbreviation for company

Explanation:

8 0

3 years ago

Read 2 more answers

What is the stress concentration factor of a shaft in torsion, where D=1.25 in. and d=1 in. and the fillet radius is, r=0.2 in.a

klio [65]

Answer:

Concentration factor will be 1.2

So option (C) will be correct answer

Explanation:

We have given outer diameter D = 1.25 in

And inner diameter d = 1 in and fillet ratio r = 0.2 in

So $\frac{r}{d}$ ratio will be $=\frac{0.2}{1}=0.2$

And $\frac{D}{d}$ ratio will be $=\frac{1.25}{1}=1.25$

Now from the graph in shaft vs torsion the value of concentration factor will be 1.2

So concentration factor will be 1.2

So option (C) will be correct answer.

6 0

3 years ago

A pin must be inserted into a collar of the same steel using an expansion fit. The coefficient of thermal expansion of the metal

nirvana33 [79]

Answer:

a) the temperature to which the pin must be cooled for assembly is $T_2 = -101.89^ \ ^0}C$

b) the radial pressure at room temperature after assembly is $P_f = 62.8 \ MPa$

c) the safety factor in the resulting assembly = 6.4

Explanation:

Coefficient of thermal expansion $\alpha = 12.3*10^{-6} \ ^0 C$

Yield strength $\sigma_y$ = 400 MPa

Modulus of elasticity (E) = 209 GPa

Room Temperature $T_1$ = 20°C

outer diameter of the collar $D_o = 95 \ mm$

inner diameter of the collar $D_i = 60 \ mm$

pin diameter $D_p$ = $60.03 \ mm$

Clearance c = 0.06 mm

a)

The temperature to which the pin must be cooled for assembly can be calculated by using the formula:

$(D_i - c )-D_p = \alpha * D_p(T_2-T_1)$

$(60-0.06)-60.03=12.3*10^{-6}*60.03(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}$ $(T_{2}-20^{0}C)$

-0.09 = $7.38369*10^{-4}T_2$ $\ \ - \ \ 0.01476738$

$-0.09 + 0.01476738$ = $7.38369*10^{-4}T_2$

−0.07523262 = $7.38369*10^{-4}T_2$

$T_2 = \frac{-0.07523262}{7.38369*10^{-4}}$

$T_2 = -101.89^ \ ^0}C$

b)

To determine the radial pressure at room temperature after assembly ;we have:

$P_f = \frac{E * (D_p-D_i)(D_o^2-D_1^2)}{D_i*D_o} \\ \\ \\ P_f = \frac{209*10^9* 0.03(95^2-60^2)}{60*95^2} \\ \\ P_f = 62815789.47 \ Pa \\ \\ P_f = 62.8 \ MPa$

c) the safety factor of the resulting assembly is calculated as:

safety factor = $\frac{Yield \ strength }{walking \ stress}$

safety factor = $\frac{400}{62.8}$

safety factor = 6.4

Thus, the safety factor in the resulting assembly = 6.4

4 0

3 years ago

A body of weight 300N is lying rough

kumpel [21]

Answer:

Horizontal force = 89.2 N

Explanation:

The frictional force = coefficient of friction * magnitude of the force (weight of the body) * cos theta

Substituting the given values, we get -

Frictional Force = 0.3*300 * cos 25 = 89.2 N

Horizontal force = 89.2 N

6 0

3 years ago