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solong [7]
2 years ago
12

What are the three major costs of owning a vehicle throughout a year?

Engineering
2 answers:
Olenka [21]2 years ago
6 0

Answer:

Owning a car is expensive. Between the price of the vehicle, financing costs, insurance, taxes, and maintenance, owning one —let alone two — cars can drain your bank account quickly.

Explanation:

ValentinkaMS [17]2 years ago
3 0

Answer:

Car Payments. Making payments on your car is the biggest, most obvious expense of your vehicle. ...

Insurance. Insurance is another primary expense to consider when budgeting for a new car. ...

Gas. ...

Maintenance. ...

Fees & Taxes.

Explanation:

You might be interested in
List two things that technological systems have in common.​
Sphinxa [80]

They all share the way that they are fundamentally designed: if they are quite complex, they will share the same basic logic foundations, like the way that the programming languages work. They also all share the method of construction and common and fundamental electronic components, like resistors, capacitors and transistors. As we humans design them, they make logical sense to at least someone, and probably only discounting the internet, you can probably draw logic diagrams and whatever to represent how they work.

Because they are designed by Humans, in a way they all mimic how our brains and society work. Also, as yet there are no truly intelligent technological systems, and are only able to react to a situation how they have been programmed to do so.

3 0
2 years ago
Anaircraft component is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 40 MPa 1/2.It has been d
navik [9.2K]

Answer:

Yes, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40MPa

Explanation:

Given

Toughness, k = 40Mpa

Stress, σ = 300Mpa

Length, l = 4mm = 4 * 10^-3m

Under which fracture occurred (i.e., σ= 300 MPa and 2a= 4.0 mm), first we solve for parameter Y (The dimensionless parameter)

Y = k/(σπ√a)

Where a = ½ of the length in metres

Y = 40/(300 * π * √(4/2 * 10^-3))

Y = 1.68 ---- Approximated

To check if fracture will occur of not; we apply the same formula.

Y = k/(σπ√a)

Then we solve for k, where

σ = 260Mpa and a = ½ * 6 * 10^-3

So,.we have

1.68 = k/(260 * π * √(6*10^-3)/2)

k = 1.68 * (260 * π * (6*10^-3)/2)

k = 42.4 MPa --- Approximately

Therefore, fracture will occur since toughness (42.4 MPa) is greater than the toughness of the material, 40 MPa

7 0
3 years ago
Read 2 more answers
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
3 years ago
Identify three operational controls and explain<br> how to use them?
ladessa [460]

Answer:

5.1 Personnel Security. ...

5.2 Physical and Environmental Protection. ...

5.3 Production, Input and Output Controls. ...

5.4 Contingency Planning and Disaster Recovery. ...

5.5 System Configuration Management Controls. ...

5.6 Data Integrity / Validation Controls. ...

5.7 Documentation. ...

5.8 Security Awareness and Training.

8 0
3 years ago
Discuss the environmental concerns regarding microfibre products
zlopas [31]

I don't know

Explanation:

cfffffffffggg

8 0
3 years ago
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