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masha68 [24]
4 years ago
12

I'm really bad at measurements so I don't understand this.

Engineering
1 answer:
cricket20 [7]4 years ago
6 0

Answer:

the answer should be 16

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Based on these statements:
Evgen [1.6K]

Answer:

the third statement is true

Explanation:

given data

Lenovos cost more than Dells

Lenovos cost less than Apples

solution

we have given 1st statement that is express as

cost (Lenovo) > cost (Dell)     ..................1

and

2nd statement that is express as

cost (Lenovo) < cost (Apple)

so we can say it as

cost (Apple) > cost (Lenovo)       ......................2

and

now above Both equation 1 and 2 can be written as

cost (Apple) > cost (Lenovo) > cost (Dell)      .........................3

so we can say cost of Apples is more than the cost of Lenovos and the cost of Dells

so as that given 3rd statement is true

7 0
4 years ago
Fully developed conditions are known to exist for water flowing through a 25-mm-diameter tube at 0.01 kg/s and 27 C. What is the
Irina18 [472]

Answer:

0.0406 m/s

Explanation:

Given:

Diameter of the tube, D = 25 mm = 0.025 m

cross-sectional area of the tube = (π/4)D² = (π/4)(0.025)² = 4.9 × 10⁻⁴ m²

Mass flow rate = 0.01 kg/s

Now,

the mass flow rate is given as:

mass flow rate = ρAV

where,

ρ is the density of the water = 1000 kg/m³

A is the area of cross-section of the pipe

V is the average velocity through the pipe

thus,

0.01 = 1000 × 4.9 × 10⁻⁴ × V

or

V = 0.0203 m/s

also,

Reynold's number, Re = \frac{VD}{\nu}

where,

ν is the kinematic viscosity of the water = 0.833 × 10⁻⁶ m²/s

thus,

Re = \frac{0.0203\times0.025}{0.833\times10^{-6}}

or

Re = 611.39 < 2000

thus,

the flow is laminar

hence,

the maximum velocity =  2 × average velocity = 2 × 0.0203 m/s

or

maximum velocity = 0.0406 m/s

5 0
3 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10
Fiesta28 [93]

Answer:

Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)

where

σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa

α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m

ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m

Substituting these values into the formula, we can calculate the max stress as

 ====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)

σₐ  = 24.4MPa

6 0
3 years ago
An urgent. Please answer this. :)<br><br>1 &amp; 2 are Ω<br><br>3 &amp; 4 are V​
balandron [24]

Answer:

See below ↓

Explanation:

1) 40 Ω

2) 24 Ω

3) 85 V

4) 135 V

6 0
3 years ago
What is the tool used to take off a wheel nut
pishuonlain [190]

Answer:

A socket wrench

Explanation:

My dad is a mechanic

8 0
3 years ago
Read 2 more answers
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