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pashok25 [27]
3 years ago
8

. A 10W light bulb connected to a series of batteries may produce a brighter lightthan a 250W light bulb connected to the same b

atteries. Why? Explain.
Engineering
2 answers:
Dvinal [7]3 years ago
3 0

Answer:

Explanation:

From the equation:

Power dissipated= square of voltage supplied by battery ÷ Resistance of the load

i.e P= V^2/R

It means that at constant voltage, the the power consumed is inversely related to the resistance. Therefore the 10W bulb which has a higher resistance will consume less power using the sufficiently excess power dissipated to glow brighter than the 250W bulb which has a low resistance. The power dissipated will partly be used to overcome this low resistance making less power available for heating up the 250W bulb .

gogolik [260]3 years ago
3 0

ANSWER:

A 10W bulb may shine brighter than a 250W bulb, when the two are connected to the same battery. This can happen when the battery is low or not fully charged. Because the 10W bulb has a high resistance more than a 250W bulb, it makes the 10W bulb to be more efficient than the 250W bulb.

When a low current is passed through the fillament of the two bulbs, the 10W bulb which has high resistance, uses almost all the current that enters the filament to emit more light, more than the 250W bulb which has a low resistance and will converts almost all the current that enters the filament into heat energy. In the 250W bulb only about 8% of the current are used to light up the bulb and the rest are converted to heat energy. This explains why a bulb which it's watt is higher will be more hotter when lighted up, than a bulb which watt is lower.

Another reason while the 10W bulb will shine brighter is when the fillament in it are coiled tight round itself ( example is a fluorescent bulb). It will emit more light than a 250W watt bulb that the filament are not coiled (example is an incandescent bulb).

NOTE : On a normal circumstances, a 250W bulb will shine brighter than a 10W bulb, because the higher the electric energy a bulb consumes the brighter light the filament will produce. Watt is the amount of electric energy the bulb can consume, therefore a bulb with 250W is assumed to produce more light than a bulb with 10W.

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What are the four categories of engineering materials used in manufacturing?
alexgriva [62]

Answer:

metals, composite, ceramics and polymers.

Explanation:

The four categories of engineering materials used in manufacturing are metals, composite, ceramics and polymers.

i) Metals: Metals are solids made up of atoms held by matrix of electrons. They are good conductors of heat and electricity, ductile and strong.

ii) Composite: This is a combination of two or more materials. They have high strength to weight ratio, stiff, low conductivity. E.g are wood, concrete.

iii) Ceramics: They are inorganic, non-metallic crystalline compounds with high hardness and strength as well as poor conductors of electricity and heat.

iv) Polymers: They  have low weight and are poor conductors of electricity and heat

8 0
2 years ago
How do engineering approach their work (how do they do to solve the problem).
Romashka-Z-Leto [24]

engineering approach their work by finding new solutions to their problems. like if two things dont go together in an invention, they will find another way

6 0
3 years ago
A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t
maksim [4K]

Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

A) The flow velocity intensity for the input radial coordinate "r" is given by;

U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

Thus,

U(r) = (V_c)[1 - (2r/D)²]

From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m

But we are to find the velocity at the centre of the tube, thus;

We will use the radius across the horizontal distance which will be;

0.05 - 0.012 = 0.038m

Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

Thus;

U(r) = (V_c)[1 - (2r/D)²]

0.8 = (V_c)[1 - {(2 * 0.038)/0.1}²]

0.8 = (V_c)[1 - (0.76)²]

V_c = 0.8/0.4224 = 1.894 m/s

B) flow rate is given by;

ΔV = Average Velocity x Area

Now, average velocity = V_c/2

Thus, average velocity = 1.894/2 = 0.947 m/s

Area(A) = πr² = π x 0.05² = 0.007854 m²

So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s

4 0
3 years ago
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lesya [120]

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2 years ago
Read 2 more answers
A car moves along a circular track of radius 250 ft such that its seed for a short period of time 0<_ t <_ 4s, is v = 3(t
beks73 [17]

Answer: a) -5 ft/s²

B) 4.5 ft

Explanation: Radius= 250ft

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A). When t= 3 s, the acceleration is

dv/dt = 1-2t

dv/dt = 1- 2(3)

= 1-6

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B. How far it traveled in 3 sec

Distance= 3t²/2 -t³/3 ft

Substituting 3

Distance = 27/2 - 9

= 13.5 - 9

= 4.5 ft

7 0
3 years ago
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