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3 years ago

11

Cell phones require powerful batteries in orde to work effectively. Which activity is best described as an engineering endeavor

related to cell phone batteries

1 answer:

natta225 [31]3 years ago

8 0

Where are the options?

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a vertical cylindrical container is being cooled in ambient air at 25 °C with no air circulation. if the initial temperature of

Sloan [31]

Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period. = 4.93 w/m²k

Explanation:

check attachement for answer explanation

7 0

3 years ago

Read 2 more answers

What is vapier calliper

SIZIF [17.4K]

Answer:

A calliper is a device used to measure the dimensions of an object.

Many types of calipers permit reading out a measurement on a ruled scale, a dial, or a digital display. Some calipers can be as simple as a compass with inward or outward-facing points, but no scale.

I'm not sure what a "Vapier" Calliper is, but I assume it's about the same thing.

4 0

2 years ago

What is the subtotal for the building materials in the bill? The subtotal should be the cost of the materials only. Don't add ta

Alexxx [7]

Answer:

Cost before adding taxes and incorporating variations

Explanation:

The sub-totals are basically sum of the cost from individual pages before including value added tax or any other tax depending with the state regulations. At this stage, contingencies and variation of prices are not considered. Therefore, the sub-total for building materials only include the materials on site or materials used as per the bill of quantities and certified by the engineer.

3 0

4 years ago

A solid shaft is subjected to an axial load P = 200 kN and a torque T = 1.5 kN.m. a) Determine the diameter of the shaft if the

Rom4ik [11]

Answer:

<em>a) 42 mm</em>

<em>b) 144.4 MPa</em>

<em></em>

Explanation:

Load P = 200 kN = 200 x 10^3 N

Torque T = 1.5 kN-m = 1.5 x 10^3 N-m

maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa

diameter of shaft d = ?

From T = τ * $\frac{\pi }{16}$ * $d^{3}$

substituting values, we have

1.5 x 10^3 = 100 x 10^6 x $\frac{3.142 }{16}$ x $d^{3}$

$d^{3}$ = 7.638 x 10^-5

d = $\sqrt[3]{7.638 * 10^-5}$ = 0.042 m = <em>42 mm</em>

b) Normal stress = P/A

where A is the area

A = $\frac{\pi d^{2} }{4}$ = $\frac{3.142*0.042^{2} }{4}$ = 1.385 x 10^-3

Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = <em>144.4 MPa</em>

7 0

4 years ago

Drag the tiles to the correct boxes to complete the pairs.

sammy [17]

Explanation:

electric motor: fan

semiconductor:

wireless technology: transistor

7 0

2 years ago