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3 years ago

1. You do not need to remove the lead weights inside tires before recycling them.

Engineering

1 answer:

makvit [3.9K]3 years ago

7 0

It true -.-.-.-.-.-.-.-.-.-.-.-.-.- EZ

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Convert 0.025 in into mm.

malfutka [58]

Answer: 0.025 in = 0.065 mm

Explanation: To convert the value in inches to mm we have to multiply the inches by the conversion factor 25.4.

So, 0.025 × 25.4 = 0.065 mm (millimeter)

Conversion formula for calculation in (inch) into mm is:

Value in mm = Value in in × 25.4

One inch is equal to the 25.4 mm.

3 0

3 years ago

Read 2 more answers

A 0.25" diameter A36 steel rivet connects two 1" wide by .25" thick 6061-T6 Al strips in a single lap shear joint. The shear str

just olya [345]

Answer:

Option B

1025 psi

Explanation:

In a single shear, the shear area is $\frac {\pi d^{2}}{4}=\frac {\pi 0.25^{2}}{4}$

The shear strength= $0.58\sigma_y$ and in this case $\sigma_y=36 000 psi$

Shear strength= $\frac {Load}{Shear area}$ hence making load the subject then

Load=Shear area X Shear strength

Load= $\frac {\pi 0.25^{2}}{4} \times 0.58\times 36000\approx 1025 psi$

3 0

3 years ago

A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

$a = -0.185\,\frac{ft}{s^{2}}$

The time required to travel is:

$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

$t = 277.477\,s\,(4.625\min)$

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be $v_{f} = 0\,\frac{mi}{h}$ .

c) The final constant deceleration is:

$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$