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3 years ago

Determine the voltage which must be applied to a 1 k 2 resistor in order that a current of

Engineering

1 answer:

sergey [27]3 years ago

5 0

Answer:

The correct solution is "20 volt".

Explanation:

Given that:

Current,

I = 10 mA

or,

= $10\times 10^{-3} \ A$

Resistance,

R = 2 K ohm

or,

= $1\times 10^3 \ ohm$

Now,

The voltage will be:

⇒ $V=IR$

By putting the values, we get

$=10\times 10^{-3}\times 2\times 10^{3}$

$= 20 \ volt$

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What is the meaning of beauty and completeness? In relation to these what are the attributes a Muslim should adopt?

disa [49]

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3 0

3 years ago

Future solution for air pollution in new zealand

andrezito [222]

Answer:

New Zealand may use some of these solutions to prevent air pollution

Explanation:

Using public transports.

Recycle and Reuse

No to plastic bags

Reduction of forest fires and smoking

Use of fans instead of Air Conditioner

Use filters for chimneys

Avoid usage of crackers

5 0

3 years ago

A sandy soil has a total unit weight of 120 pcf, a specific gravity of solids of 2.64, and a water content of 16 percent. Comput

olchik [2.2K]

Answer:

A). Dry unit weight = 1657.08Kg/m3

B). Porosity = 0.37

C). Void ratio = 0.593

D). 0.712

Explanation:

Total unit weight, Y = 120pcf =1922.2 Kg/m3

Specific gravity of solids, Gs = 2.64

Water content, w = 16%

A). Dry unit weight

Yd = Y/(1+w)

= 1922.2/(1+0.16) = 1657.08Kg/m3

B). Porosity

However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3

Void ratio = 2.64×1000/1657.08 = 0.593

And porosity = e/(1+e) =0.593/(1+0.593) = 0.37

C). void ratio, e = 0.593

D). Degree of saturation, S = m×Gs/e where m =water content

S = 0.16×2.64/0.593 = 0.712

5 0

4 years ago

A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected

Pavlova-9 [17]

Answer:

the required diameter of the rod is 9.77 mm

Explanation:

Given:

Length = 1.5 m

Tension(P) = 3 kN = 3 × 10³ N

Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa

E = 70 GPa = 70 × 10⁹ Pa

δ = 1 mm = 1 × 10⁻³ m

The required diameter(d) = ?

a) for stress

The stress equation is given by:

$S = \frac{P}{A}$

A is the area = πd²/4 = (3.14 × d²)/4

$S = \frac{P}{(\frac{3.14*d^{2} }{4}) }$

$S = \frac{4P}{{3.14*d^{2} } }$

$3.14*S*{d^{2}} = {4P}$

${d^{2}} =\frac{4P}{3.14*S}$

$d= \sqrt{\frac{4P}{3.14*S} }$

Substituting the values, we get

$d= \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }$

$d= \sqrt{\frac{12000 }{125600000 } }$

$d= \sqrt{9.55*10^{-5} }$

d = (9.77 × 10⁻³) m

d = 9.77 mm

b) for deformation

δ = (P×L) / (A×E)

A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063

d² = (4 × A) / π = (0.000063 × 4) / 3.14

d² = 0.0000819

d = 9.05 × 10⁻³ m = 9.05 mm

We use the larger value of diameter = 9.77 mm

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3 years ago

Chemical manufacturers must present which information on the product's label?

Wewaii [24]

Answer:

hazardous chemicals leaving the workplace is labeled, tagged or marked with the following information: product identifier; signal word; hazard statement

Explanation:

this is so you know what chemicals are in it

6 0

4 years ago