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2 years ago

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]

Take a look at the pictures that should help you out.

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]

Answer:

iऑटठठषठड

Explanation:

ಓಠಪೌಡಠಚಪಠ

ghhhfyvy

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