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3 years ago

Compare and contrast fluorescent and incandescent light bulbs.

1 answer:

5 0

Incandescent light is a glowing white light produced by heat. An incandescent light bulb works by heating a filament in the bulb. Fluorescent light is a bright light produced by electricity flowing through a tube filled with ionized gas. Fluorescent light bulbs are more energy-efficient than incandescent bulbs

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Light travels 186 000 miles per second.how many miles dose light travel in one year

Troyanec [42]

(186,000 mi/sec) x (3,600 sec/hr) x (24 hr/da) x (365 da/yr)

= (186,000 x 3,600 x 24 x 365) mi/yr

= 5,865,696,000,000 miles per year (rounded to the nearest million miles)

8 0

3 years ago

Suppose you have two solid bars, both with square cross-sections of 1 cm2. They are both 24.6 cm long, but one is made of copper

vodka [1.7K]

Explanation:

Expression to calculate thermal resistance for iron ( $R_{I}$ ) is as follows.

$R_{I} = \frac{L_{I}}{k_{I} \times A_{I}}$

where, $L_{I}$ = length of the iron bar

$k_{I}$ = thermal conductivity of iron

$A_{I}$ = Area of cross-section for the iron bar

Thermal resistance for copper ( $R_{c}$ ) = \frac{L_{c}}{k_{c} \times A_{c}}[/tex]

where, $L_{c}$ = length of copper bar

$k_{c}$ = thermal conductivity of copper

$A_{c}$ = Area of cross-section for the copper bar

Now, expression for the transfer of heat per unit cell is as follows.

Q = $\frac{(100^{o} - 0^{o}}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

Putting the given values into the above formula as follows.

Q = $\frac{(100^{o} - 0^{o})}{\frac{L_{I}}{k_{I}.A_{I}} + \frac{L_{c}}{k_{c}.A_{c}}}$

= $\frac{(100^{o} - 0^{o})}{21 \times 10^{-2} m[\frac{1}{73 \times 10^{-4}m^{2}} + \frac{1}{386 \times 10^{-4}m^{2}}}$

= 2.92 Joule

It is known that heat transfer per unit time is equal to the power conducted through the rod. Hence,

P = $\frac{Q}{T}$

Here, T is 1 second so, power conducted is equal to heat transferred.

So, P = 2.92 watt

Thus, we can conclude that 2.92 watt power will be conducted through the rod when it reaches steady state.

7 0

3 years ago

Which of the following is true about conduction? Question 2 options: It is heat transfer through direct contact It can only occu

kramer

Answer:

through direct contact

Explanation:

took the test!

3 0

3 years ago

Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum

Alina [70]

Answer:

Uncertainty in position (∆x) = 24 × 10⁻¹⁵ m
Uncertainty in momentum (∆P) = ?
Planck's constant (h) = 6.26 × 10⁻³⁴ Js

$\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi}$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} }$

$\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} }$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192}$

$\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192}$

$\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21}$

$\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21}$

$\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}$

4 0

3 years ago

If we keep on applying force on a material object, can it gain speed of light?

Xelga [282]

Answer:

As an object moves faster, its mass increases. ... Because masses approach infinity with increasing speed, it is impossible to accelerate a material object to (or past) the speed of light. To do so would require an infinite force.

8 0

3 years ago