All categories

3 years ago

Fast plz-The mirror check may involve ______________.

Engineering

1 answer:

barxatty [35]3 years ago

6 0

Answer:

Realigning the mirror

Explanation:

mirrors should be aligned to minimize blind spots, not look at the tires.

You might be interested in

Energy is generated uniformly in a 6 cm thick wall. The steady-state temperature distribution

mezya [45]

Answer:

Heat generation per unit volume is 45 KW.

Explanation:

Given that

Thickness of wall = 6 cm

Temperature distribution

$T(z)=145+3000z-1500z^2$ -----1

K= 15 W/m.k

As we know that at steady state condition

$\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0$ -----2

Where q is the heat generation per unit volume.

So from equation 1

$\dfrac{dT}{dz}=3000-3000z$

$\dfrac{d^2T}{dz^2}=-3000$

Now from equation 2

$\dfrac{d^2T}{dz^2}+\dfrac{q}{K}=0$

$-3000+\dfrac{q}{15}=0$

So q= 45 KW

So heat generation per unit volume is 45 KW.

6 0

3 years ago

A cylindrical hot water storage tank (see sketch) for a set of collectors is located in the basement of a dwelling. The tank is

Galina-37 [17]

Answer:

See explanations

Explanation:

Given Data, The diameter of the Tank (D) = 2 ft.

Height of the Tank (H) = 4.5 ft.

Inside temperature of Tank (T1) = 120° F Outside temperature (T2) = 65° F The thickness of Steel (s1) = 0.080" = 0.00666667 ft. Thickness of fiberglass

8 0

3 years ago

A homogeneous 800kg bar AB is supported at either end by a cable asshown in the figure

aleksandr82 [10.1K]

The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².

<h3>What is normal stress?</h3>

If the direction of deformation force is perpendicular to the cross-sectional area of the body, the stress is called normal stress. Changes in wire length and body volume will be normal.

σ = P/A

Where, σ = Normal stress

P = Pressure

A = Area

1 Kg = 9.81 N

800 kg = 7848 N

Since the rod is half bronze and half steel

800 kg = 7848/2

= 3924 N

Pₙ = Fₙ = 3924 N [n = Bronze]

Pₓ = 3924 N [x = steel]

Given,

σₙ = 90MPa

σₓ = 120MPa

Aₙ = ?

Aₓ = ?

Aₙ = Pₙ/σₙ

Aₙ = 3924/90

Aₙ = 43.6 mm²

Aₓ = Pₓ/σₓ

Aₓ = 3924/120

Aₓ = 32.7 mm²

To know more about normal stress, visit:

brainly.com/question/28012990

#SPJ9

4 0

1 year ago

Motorcycles are extremely hard to see if they are _______. powered by quiet motors approaching from the side driving on the shou

Assoli18 [71]

Answer: uve earned 5 b point for helping us

Explanation:well know I'm not helping u

7 0

3 years ago

Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a

guapka [62]

Answer:

$K_v=12.34$

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo = Discharge / area of the orifice

Vo = $\frac{0.01}{\frac{\pi}{4}0.045^2}$

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, $V_v$ = Discharge / area of the orifice

$V_v$ = $\frac{0.01}{\frac{\pi}{4}0.0675^2}$

$V_v$ = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

25 m = $K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}$

where,

$K_v$ is the loss coefficient for the gate valve

on substituting the values, we get

25 m = $10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}$

$K_v\frac{2.79^2}{2\times9.81}$ = 4.898

$K_v=12.34$

3 0

4 years ago