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2 years ago

8

1. Which type of fit implies that a piece will never fit? a. interference fit b. construction fit c. transition fit d. impeding

fit

1 answer:

VladimirAG [237]2 years ago

6 0

Answer:

d

Explanation:

interference fit: the dimensions are larger and high force like heating is required to fit in the object

transition fit: the dimensions are fractionally larger and slight force like hammering is required to fit in the object

impeding fit: the dimensions are larger and the object will not fit at all

construction fit: activities making a commercial space sutiable for tenant occupation.

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Which of the following activities can help expand engineers' creative thinking capabilities?

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2 years ago

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The volume of the pyramid is 36 cubic cm, find the volume of the prism.

ser-zykov [4K]

Answer:

Given :- the volume of the pyramid is 36 cubic cm , find the volume of the prism on same base and same height as pyramid .

Answer :-

we know that,

Volume of pyramid = (1/3) * Base area * height .

Volume of prism = Base area * height .

so,

→ Volume of pyramid = 36 cm³

→ (1/3) * Base area * height = 36

→ Base area * height = 36 * 3

→ Base area * height = 108 cm³.

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Explanation:

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2 years ago

The first step to merging is entering the ramp and _____.

Yuri [45]

Answer:

D.telling your passengers where you are going

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3 years ago

The car travels around the portion of a circular track having a radius of r = 500 ft such that when it is at point A it has a ve

stellarik [79]

Answer:

Explanation:

Given

velocity at A is $v=4\ ft/s$

For $r=500\ ft$

velocity is increasing at $\dot{v}=0.004t\ ft/s^2$

Tangential acceleration is given by

$a_t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a_t=0.004t=\frac{\mathrm{d} v}{\mathrm{d} t}$

$\int 0.004tdt=\int dv$

$\int dv=\int 0.004tdt$

$v=0.002t^2+c$

at $t=0\ v=4\ ft/s$

$4=0.002\cdot 0+c$

$c=4\ ft/s$

thus $v=0.002t^2+4$

Velocity in terms of Displacement is given by

$v=\frac{\mathrm{d} s}{\mathrm{d} t}$

$\Rightarrow \int ds=\int \left ( 0.002t^2+4\right )dt$

$\Rightarrow s=\frac{0.002t^3}{3}+4t$

When car has traveled $\frac{3}{4}$ th of distance i.e.

$s=\frac{3}{4}\times (2\pi r)=\frac{3\pi r}{2}$

$s=750\pi$

$750\pi =\frac{0.002t^3}{3}+4t$

$\Rightarrow \frac{0.002t^3}{3}+4t-2356.5=0$

on solving we get $t=139.23\ s$

Thus velocity at $t=139.23\ s$

$v=42.76\ s$

(b)Acceleration when car has traveled three-fourth the way of track

normal acceleration $a_n=\frac{v^2}{r}=\frac{(42.76)^2}{500}$

$a_n=3.658\ m/s^2$

Tangential acceleration $a_t at t=139.23\ s$

$a_t=0.556\ m/s^2$

Net acceleration $a_t=\sqrt{(a_n)^2+(a_t)^2}$

$a_n=\sqrt{(3.658)^2+(0.556)^2}$

$a_n=3.7\ m/s^2$

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3 years ago

Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the r

9966 [12]

Answer:

(a) Calculate the rod base temperature (°C). = 299.86°C

(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m

Explanation:

see attached file below

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3 years ago