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3 years ago

What is the advantage of using fluorescent bulbs over incandescent bulbs?

1 answer:

5 0

Incandescent bulbs work by conducting an electric current along a filament made of a long, thin piece of tungsten metal. The filament must be heated to temperatures of about 2,300 degrees Celsius to glow and emit a white-hot light. But the process transforms only 5 percent to 10 percent of the electricity used into visible light. The rest is transformed into heat, which can eventually increase the temperature of a room.

CFL bulbs, on the other hand, are made of glass tubes filled with gas and a small amount of mercury. The amount is so small that an old-fashioned glass thermometer holds 100 times as much mercury as one CFL bulb. Light is emitted when mercury molecules in a CLF bulb become excited by electricity running between two electrodes at its base. The mercury emits an invisible ultraviolet light that becomes visible when it hits the white coating inside the CFL bulb.

Basically, based on how they both work, fluorescent bulbs last longer and do not release as much heat as incandescent bulbs which means it's also better for the environment.

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You don't have any matches, so you want to focus sunlight to ignite a pile of twigs for a campfire. Which mirror would be best t

Anna35 [415]

Out of that list, the concave mirror is the only item that can concentrate sunlight and heat into a small area. But if you could get ahold of a convex lens, that would be even better.

4 0

4 years ago

Read 2 more answers

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju

denpristay [2]

Answer:

a) $h=250\ m$

b) $\Delta h=0.0835\ m$

Explanation:

Given:

upward acceleration of the helicopter, $a=5\ m.s^{-2}$

time after the takeoff after which the engine is shut off, $t_a=10\ s$

Maximum height reached by the helicopter:

using the equation of motion,

$h=u.t+\frac{1}{2} a.t^2$

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

$h=0+0.5\times 5\times 10^2$

$h=250\ m$

time after which Austin Powers deploys parachute(time of free fall), $t_f=7\ s$

acceleration after deploying the parachute, $a_p=2\ m.s^{-2}$

height fallen freely by Austin:

$h_f=u.t_f+\frac{1}{2} g.t_f^2$

where:

$u=$ initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

$t_f=$ time of free fall

$h_f=0+0.5\times 9.8\times 7^2$

$h_f=240.1\ m$

Velocity just before opening the parachute:

$v_f=u+g.t_f$

$v_f=0+9.8\times 7$

$v_f=68.6\ m.s^{-1}$

Time taken by the helicopter to fall:

$h=u.t_h+\frac{1}{2} g.t_h^2$

where:

$u=$ initial velocity of the helicopter just before it begins falling freely = 0

$t_h=$ time taken by the helicopter to fall on ground

$h=$ height from where it falls = 250 m

now,

$250=0+0.5\times 9.8\times t_h^2$

$t_h=7.1429\ s$

From the above time 7 seconds are taken for free fall and the remaining time to fall with parachute.

remaining time,

$t'=t_h-t_f$

$t'=7.1428-7$

$t'=0.1428\ s$

Now the height fallen in the remaining time using parachute:

$h'=v_f.t'+\frac{1}{2} a_p.t'^2$

$h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2$

$h'=9.8165\ m$

Now the height of Austin above the ground when the helicopter crashed on the ground:

$\Delta h=h-(h_f+h')$

$\Delta h=250-(240.1+9.8165)$

$\Delta h=0.0835\ m$

5 0

3 years ago

Tyler's favourite number is 8642, and he tries to fit it into everything he does. If he wishes to do that much work by climbing

Brums [2.3K]

Answer:

2014

Explanation:

none

6 0

3 years ago

A test car starts from rest on a horizontal circular track of 115-m radius and increases its speed at a uniform rate to reach 90

Wewaii [24]

Answer:

a= 3.49 m/s^2

Explanation:

magnitude of total acceleration = sqrt{radial acceleration^2+tangential acceleration^2}.

we know that tangential acceleration a_t= change in velocity /time taken

now 90 km/h = 25 m/s

a_t = 25/17 = 1.47 m/s^2.

radial acceleration a_r = v^2/r

v= a_t×t = 1.47×13 = 19.11 m/s

a_r = 19.11^2/115= 3.175

now,

$a= \sqrt{a_t^2+a_r^2}$

$a= \sqrt{1.47^2+3.175^2}$

a= 3.49 m/s^2

3 0

3 years ago

Light does not pass through some materials. What do you think happens

Hatshy [7]

Answer:

When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. One way of thinking about this higher energy state is to imagine that the electron is now moving faster, (it has just been "hit" by a rapidly moving photon)

A photon is a quantum of EM radiation. Its energy is given by E = hf and is related to the frequency f and wavelength λ of the radiation by. E=hf=hcλ(energy of a photon) E = h f = h c λ (energy of a photon) , where E is the energy of a single photon and c is the speed of light.

7 0

2 years ago