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pashok25 [27]
4 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]4 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]4 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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Pendulum impacting an inclined surface of a block attached to a spring-Dependent multi-part problem assign all parts NOTE: This
Art [367]

Answer:

vA = -2.55 m/s

vB = 0.947 m/s

Explanation:

Given:-

- The initial angle of rope, α = 30°

- The angle of rope just before impact or wedge angle, θ = 20°

- The weight of sphere, Ws = 1-lb

- The initial position velocity, vi = 4 ft/s

- The coefficient of restitution, e = 0.7

- The weight of the wedge, Ww = 2-lb

- The spring constant, k = 1.8 lb/in

- The length of rope, L = 2.6 ft

Find:-

 Determine the velocities of A and B immediately after the impact.

Solution:-

- We can first consider the ball ( acting as a pendulum ) to be isolated for study.

- There are no unbalanced fictitious forces acting on the sphere ball. Hence, we can reasonably assume that the energy is conserved.

- According to the principle of conservation for the initial point and the point just before impact.

Let,

              vA : The speed of sphere ball before impact

               

                  Change in kinetic energy = Change in potential energy

                  ΔK.E = ΔE.P

                  0.5*ms* ( uA^2 - vi^2 ) = ms*g*L*( cos ( θ ) - cos ( α ) )

                  uA^2 = 2*g*L*( cos ( θ ) - cos ( α ) ) + vi^2

                  uA = √ [ 2*32*2.6*( cos ( 20 ) - cos ( 30 ) ) + 4^2 ] = √28.25822

                  uA = 5.316 ft/s

- The coefficient of restitution (e) can be thought of as a measure of the extent to which mechanical energy is conserved when an object bounces off a surface:

                 e^2 = ( K.E_after impact / K.E_before impact )

- The respective Kinetic energies are:

               

                K.E_after impact = K.E_sphere + K.E_block

                                             = 0.5*ms*vA^2 + 0.5*mb*vB^2

                K.E_before impact = K.E = Ws*L*( cos ( θ ) - cos ( α ) )

                                                         = 1*2.6*( cos ( 20 ) - cos ( 30 ) )

                                                         = 0.1915 J

                32*2*0.1915*0.7^2 = Ws*vA^2 + Wb*vB^2  

                6.00544 = vA^2 + 2*vB^2  ... Eq1

- From conservation of linear momentum we have:

                vB = e*( uA - uB )*cos ( 20 ) + vA

                vB = 0.7*( 5.316 - 0 )*cos ( 20)   + vA

                vB = 3.49678 + vA  .... Eq 2

- Solve two equation simultaneously:

               

               6.00544 = vA^2 + 2*(3.49678 + vA)^2

               6.00544 = 3vA^2 + 13.98*vA + 24.455

               3vA^2 + 14.8848*vA + 18.4495 = 0

               vA = -2.55 m/s

               vB = 0.947 m/s

                                 

5 0
4 years ago
<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%20%7B%5Cfrac%7Bt-1%7D%7B1-t%5E3%7D%20%7D%20%5C%2C%20dt" id="TexFormula1" title
Flura [38]

Answer:

-2/√3 atan ((2t + 1)/√3) + C

Explanation:

∫ (t − 1) / (1 − t³) dt

Factor the difference of cubes:

∫ (t − 1) / ((1 − t)(1 + t + t²)) dt

Divide:

∫ -1 / (1 + t + t²) dt

-∫ 1 / (t² + t + 1) dt

Complete the square:

-∫ 1 / (t² + t + ¼ + ¾) dt

-∫ 4 / (4t² + 4t + 1 + 3) dt

-∫ 4 / ((2t + 1)² + 3) dt

If u = 2t + 1, du = 2 dt:

-∫ 2 / (u² + 3) du

Use an integral table, or use trigonometric substitution:

-2 (1/√3) atan (u/√3) + C

-2/√3 atan (u/√3) + C

Substitute back:

-2/√3 atan ((2t + 1)/√3) + C

4 0
3 years ago
The use of seatbelts in a car has significantly reduced the number of crash fatalities. Which statement best explains how societ
satela [25.4K]

as the public need for preventing injuries and deaths from car crashes became known,laws were enacted to mandate the inclusion of seatbelts in almost all passengers vehicles

8 0
3 years ago
An electric water heater held at 120° F is kept in a 60°F room. When purchased, its insulation is equivalent to R-5. An owner pu
Trava [24]

Answer:

Total saving would be of 36.917 $\yr

Explanation:

Given Data:

T_{heater} = 120 Degree F

T_{room} = 60 Degree F

A = 30 ft^2

\eta = 100%

Heat loss before previous final value = \frac{A \Delta T}{R}

                                                              =\frac{30\times *(120-60)}{5}

                                                              = 360 Btu/hr

Heat loss after new value= \frac{30\times \times (120-60)}{15} = 120 Btu/hr

saving would be = 360 - 120 Btu/hr \times kw hr/ 3412 Btu\times 24 hr/day \times 365 day/year

                           = 616.1782 kw hr/yr

cost = 616.1782 \times 0.06$

         = 36.917 $\yr

6 0
4 years ago
Please help
Alla [95]

Answer:

Explanation:

?

5 0
3 years ago
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