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Dafna11 [192]
3 years ago
15

To solve the problem, make assumptions for missing data and justify. Given:

Engineering
1 answer:
finlep [7]3 years ago
6 0

Answer:

5,4,1, this is a explication

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Who plays warzone ps4?
dsp73
Who doesn’t play war zone
7 0
3 years ago
Read 2 more answers
A market structure in which ther a single buyer
faltersainse [42]

Answer:

Monopsony

Explanation:

Kinda like a monopoly, except instead of one producer it is one buyer.

8 0
2 years ago
forty gal/min of a hydrocarbon fuel having a spesific gravity of 0.91 flow into a tank truck with load limit of 40,000 lb of fue
Zina [86]

Answer: 131.75minutes

Explanation:

First if all, we've to find the density of liquid which will be:

= Specific gravity × Density to pure water

= 0.91 × 8.34lb/gallon

= 7.59lb/gallon

Then, the volume that's required to fill the tank will be:

= Load limit/Density of fluid

= 40000/7.59

= 5270.1gallon

Now, the time taken will be:

= V/F

= 5270.1/40

= 131.75min

It'll take 131.75 minutes to fill the tank in the truck.

5 0
3 years ago
I need answer to this question
blagie [28]

Answer:

1. Graph C

2. Friction

Explanation:

1. The line on all of the graphs shown represents velocity. The formula for velocity is v=\frac{d}{t} where d is distance and t is time. Focusing on the first lap, the starting point on the graph should be the origin and the "ending" point should be (20, 3). These requirements eliminate graph A as an answer because its "end" is not (20, 3). During the break, the student does not move, so the slope of the line should be completely horizontal. The break lasted for 5 minutes, so the correct graph should have a horizontal line between the points (20, 3) and (25, 3). This requirement eliminates graph B and D because their break is either not long enough (B) or too long (D).

2. Friction slows down the movement of objects. When an object is rough, it produces more friction which causes the object to be slowed more. When an object is smooth, friction slows it less than it would for a rough object.

6 0
3 years ago
Neglecting the presence of friction, air drag, and other inefficiencies, how much gasoline is consumed when a 1300 kg automobile
koban [17]

Answer:

Explanation:

Given that, .

Mass of car is

M = 1300kg

Velocity of car

V = 80km/h = 80 × 1000/3600

V = 22.22m/s

Calculate the kinetic energy of the vehicle as follows:

K.E = ½ MV²

K.E = ½ × 1300 × 22.22²

K.E = 320,987.65 J

Given that,

Enthalpy is 45MJ / kg

h = 45MJ / kg

Then, enthalpy is given as.

Enthalpy = Energy / mass

h = E / m

45 × 10^6 = 320,987.65 / m

m = 320,987.65 / 45 × 10^6

m = 7.133 × 10^-3 kg

m = 7.133 mg

Also, given that, density is 680kg/m³

Density is given as

Density = mass / Volume

ρ = m / v

Then, v = m / ρ

v = 7.133 × 10^-3 / 680

v = 1.049 × 10^-5 m³

We know that

1mL = 10^-6 m³

Therefore,

v = 1.049 × 10^-5 m³ × 1mL / 10^-6m³

v = 10.49 mL

7 0
3 years ago
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