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Bogdan [553]
3 years ago
10

A set of civil construction plans shows a distance of 131 meters across the front of a piece of property. You need to convert th

is value to feet. You know that one foot equals approximately 0.305 meters. What is the distance in​ feet, to the nearest tenth of a​ foot?
Engineering
1 answer:
Hoochie [10]3 years ago
7 0

Answer:

429.5 ft

Explanation:

We are told that;

1 ft = 0.305 meters

We want to convert 131 meters to ft, by proportion we have;

Distance = (131 × 1)/0.305

Distance = 429.5082 ft

We want to approximate to the nearest tenth of a foot.

This gives us;

Distance = 429.5 ft

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no artical shoul be used here

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2 years ago
Who ate cdonalds in hell with god then died from food ppoisoning
siniylev [52]

Answer:

McDonald’s announced recently that they are going through some major menu changes, and will be nixing some unnecessary ingredients. They also are finally listening to us, and will stop using chickens that are injected with growth-promoting antibiotics, along with dairy products raised with the growth hormone rbST but they still are using a lot of factory farmed meat and the beef is still raised with antibiotics.

McDonald’s even said they might add kale to their menu, by putting it in salads or in a smoothie. I LOVE kale, and I hope they don’t find a way to ruin it. You know the saying, “You can put lipstick on a pig, but it’s still a pig”. So, they need to do a lot more than just add kale to their menu before I’d ever eat there. The problem is that millions are still eating there and consuming several questionable food additives that McDonald’s could remove entirely if they really wanted to.

Explanation:

7 0
2 years ago
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve
lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0
2 years ago
Engine oil (unused) flows at 1.81 x 10^-3 kg/s inside a 1-cm diameter tube that is heated electrically at a rate of 76 W/m. At a
Ann [662]

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

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Answer:

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Explanation:

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