On a Test please help
When towing a trailer, you should consider all of the following except

the minimum speed limit on roadways.
your personal protective equipment.
the gross weight of the load.
the increased stopping distance

Answer:

HUHHHHHH BE SPECIFIC CHILE

Explanation:

ERM IRDK SORRY BOUT THAT

Answer:

the answer is below

Explanation:

a) The conductivity of graphite (σ) is calculated using the formula:

[tex]\delta=\frac{1}{\sqrt{\pi f \mu \sigma} }\\\\\sigma =\frac{1}{\pi f \mu \delta^2}[/tex]

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012

Substituting:

[tex]\sigma =\frac{1}{\pi *10^6* 0.0000012*0.00016^2}=0.99*10^4\ S/m[/tex]

b) f = 1 GHz = 10⁹ Hz.

[tex]\alpha=\sqrt{\pi f \mu \sigma} = \sqrt{0.0000012*10^9*\pi*0.99*10^5}=1.98*10^4\ Np/m\\\\20log_{10} e^{-\alpha z}=-30\ dB\\\\(-\alpha z)log_{10} e=-1.5 \\\\z=\frac{-1.5}{log_{10} e*-\alpha} =1.75*10^{-4}\ m=0.175\ mm[/tex]

Answer:

a)

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b) 3.1 secs

Explanation:

a) Determine the normal times in TMUs for these motion elements

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b ) Determine the total time for this work element in seconds

first we have to determine the total TMU = ∑ TMU = 86.4 TMU

note ; 1 TMU = 0.036 seconds

hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds

Answer: B. 15 feet

Explanation:

Answer: the allowable load P is 242.7877 kips

Explanation:

Given that;

diameter bolts d = 1.83 in

ultimate shear strength of the bolts = 60 ksi

we know that

shear area = 2×(π/4)d²

= 2×(π/4)×(1.83)² = 5.2604 in²

so

p/3(5.2604) = 60000/3.9

p/15.7812 = 15384.6153

p = 15.7812 × 15384.6153

p = 242787.691 lb

p = 242.7877 kips

therefore the allowable load P is 242.7877 kips

Answer:

[tex]95.914\ \text{GJ}[/tex]

[tex]\$272.78[/tex]

Explanation:

m = Mass of water = 749511.5 kg

c = Specific heat of water = 4182 J/kg ⋅°C

[tex]\Delta T[/tex] = Change in temperature = [tex]80.6-50=30.6^{\circ}\text{F}[/tex]

Cost of 1 GJ of energy = $2.844

Heat required is given by

[tex]Q=mc\Delta T\\\Rightarrow Q=749511.5\times 4182\times 30.6\\\Rightarrow Q=95.914\times 10^9\ \text{J}=95.914\ \text{GJ}[/tex]

Amount of heat required to heat the water is [tex]95.914\ \text{GJ}[/tex].

Cost of heating the water is

[tex]95.914\times 2.844=\$272.78[/tex]

Cost of heating the water to the required temperature is [tex]\$272.78[/tex].

Answer:

0.4 gallons per second

Explanation:

A function shows the relationship between an independent variable and a dependent variable.

The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.

The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).

Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.

Constant of proportionality = g / t = 2 / 5 = 0.4 gallons per second

Answer:

c) site preparation

Explanation:

A construction process can be defined as a series of important physical events (processes) that must be accomplished during the execution of a construction project.

Generally, in the construction of any physical asset such as offices, hospitals, schools, stadiums etc, the first step of the construction process is site preparation. Site preparation refers to processes such as clearing, blasting, levelling, landfilling, surveying, cutting, excavating and demolition of all unwanted objects on a piece of land, so as to make it ready for use.

This ultimately implies that, site preparation should be the first task to be accomplished in the construction process.

Hence, the construction process typically begins with site preparation before other activities such as the laying of foundation can be done.

Additionally, construction costs can be defined as the overall costs associated with the development of a built asset, project or property. The construction costs is classified into two (2) main categories and these are; capital and operational costs.

Solution :

Taken moment about point B;

[tex]$\sum M_B = 0$[/tex]

[tex]$\left[(w)(4)(2)-N_A \sin 30^\circ (3)(\sin 30^\circ)-N_A \cos 30^\circ(3 \cos 30^\circ +4) \right] =0$[/tex]

[tex]$N_A = 1.2376w$[/tex]

[tex]$\sum F_x = 0$[/tex]

[tex]$N_A \sin 30^\circ - B_x = 0$[/tex]

[tex]$B_x = 1.2376w \times \sin 30^\circ$[/tex]

[tex]$B_x = 0.6188w$[/tex]

[tex]$\sum F_y = 0$[/tex]

[tex]$B_y+N_A \cos^\circ - (w \times 4) = 0$[/tex]

[tex]$B_y+(1.2376w) \cos 30^\circ - (w \times 4)=0 $[/tex]

[tex]$B_y = 2.9282w$[/tex]

Now calculating the resultant force at B,

[tex]$F_B= \sqrt{B_x^2+B_y^2}$[/tex]

[tex]$F_B= \sqrt{(0.6188w)^2+(2.9282w)^2}$[/tex]

     = 2.9929w

For no failure,

[tex]$N_A<4 \ kN $[/tex]

2.9929 w < 4 kN

[tex]$w < 3.232 \ kN/m$[/tex]

And,

[tex]$F_B < 8 \ kN$[/tex]

2.9929 w < 8 kN

w < 2.673 kN/m

For the safe operation, w = 2.673 kN/m  

Answer:

I got iron

Explanation:

on my plato test