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RideAnS [48]
2 years ago
6

Q1: The first option should always be to get out safely (RUN)

Engineering
1 answer:
nekit [7.7K]2 years ago
6 0

Answer:

Q1 true

Q2 true

And other I am confuse

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Hydrogen peroxide, H2O2, enters a gas generator at 25 Celsius, 500 kPa, at the rate of 0.1 kg/s and is decomposed to steam and o
Blizzard [7]

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8 0
3 years ago
Consider a cylindrical nickel wire 2.1 mm in diameter and 3.2 × 104 mm long. Calculate its elongation when a load of 280 N is ap
pshichka [43]

Answer:

Total elongation will be 0.012 m

Explanation:

We have given diameter of the cylinder = 2.1 mm

Length of wire L=3.2\times 10^4mm

So radius r=\frac{d}{2}=\frac{2.1}{2}=1.05mm=1.05\times 10^{-3}m

Load F = 280 N

Elastic modulus = 207 Gpa

Area of cross section A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2

We know that elongation in wire is given by \delta =\frac{FL}{AE}, here F is load, L is length, A is area and E is elastic modulus

So \delta =\frac{FL}{AE}=\frac{280\times 32}{3.461\times 10^{-6}\times 207\times 10^9}=0.012m

4 0
3 years ago
Unless otherwise posted, the maximum speed limit is ______mph on a two-lane undivided highways and for vehicles towing trailers.
Karo-lina-s [1.5K]

Answer:

55 mph

Explanation:

6 0
3 years ago
Design a product counter using microcontroller
elena-s [515]

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ಓಠಪೌಡಠಚಪಠ

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8 0
3 years ago
A cylindrical bar of metal having a diameter of 20.3 mm and a length of 205 mm is deformed elastically in tension with a force o
Zarrin [17]

Answer:

a) 0.4393 mm

b)  -0.141 mm

Explanation:

Cylindrical bar :  Diameter = 20.3 mm , Length = 205 mm

Force that deforms bar of metal = 46400 N

elastic modulus = 66.9 GPa

Poisson's ratio ( u ) = 0.32

A) Determine the amount by which this specimen will elongate in the direction of applied stress

dl = \frac{P*L}{A*E}

where P = 46400 N

          L = 205 mm

          A = \frac{\pi }{4} * 20.3^2 = 323.65

          E = 66.9 GPa

dl =  ( 46400 * 205 ) / ( 323.65 * 66.9 * 10^3 )  = 0.4393 mm

B) determine the change in diameter of the specimen

change in diameter( compressed due to elongation in length )

= - u * dl

 = - 0.32 * 0.4393

=  -0.141 mm

5 0
3 years ago
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