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swat32
3 years ago
10

True or False: The differential lock in an AWD-equipped vehicle can be used at any time.

Engineering
2 answers:
Bingel [31]3 years ago
7 0

the answer would be false

Tamiku [17]3 years ago
6 0

Answer:true

Explanation:becuase yes

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What type of degree is needed to be an architect?
Rina8888 [55]
Architects must have a professional bachelor's or master's degree in architecture from a program that has been accredited by the National Architectural Accrediting Board, and a state license.
3 0
3 years ago
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A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball
faltersainse [42]

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

\sigma = \frac{p*r}{2t}

<u>Where:</u>

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm  

Hence, we need to find r, as follows:

r_{inner} = r_{outer} - t    

r_{inner} = \frac{d}{2} - t

<u>Where:</u>

d: is the outer diameter = 300 mm

r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm

Now, we can find the normal stress (σ) in the wall of the basketball:

\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!

3 0
3 years ago
Identify how slip affects the strength of a metal​
Fiesta28 [93]

My answer says there is a link but not,there isn’t so here a photo of my answer.

5 0
3 years ago
A 5 mm dia copper cable is insulated with a material of conductivity of 0.16 W/mK and is exposed to air at 30°C with a convecti
Kipish [7]

Answer:

t=5.5 mm

Heat dissipation per unit length = 90.477 W/m

Explanation:

Given that

Diameter d = 5 mm ⇒r = 2.5 mm

Conductivity of insulated material K = 0.16 W/mK

Heat transfer coefficient = 20 \frac{W}{m^2K}

When thickness reaches up to critical radius of insulation then heat dissipation will be maximum

We know that critical radius of insulation of wire is given as follow

r_{c}=\dfrac{K_{insulation}}{h_{surrounding}}

Now by putting the values

r_{c}=\dfrac{0.16}{20}

r_{c}=8 mm

So the thickness of insulation

t=8-2.5 mm

t=5.5 mm

As we know that heat transfer due to convection given as follows

Q = hAΔ T

Q=20 x 2 x π x 0.008 x (120-30)

Q = 90.477 W/m

So heat dissipation per unit length = 90.477 W/m

4 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0
3 years ago
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