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4 years ago

Which element is used in light bulbs as a fire?

Physics

2 answers:

DochEvi [55]4 years ago

5 0

The filament used in light bulb is Tungsten(W).. It's a metal with melting point of about 3300 °C.

Kitty [74]4 years ago

4 0

"Fire" is exactly what you DON't want inside a light bulb. If fire was
burning in there, it would burn up the tiny wire that glows to make the
light, and then there would be no more light from the bulb.

Thomas Edison's biggest accomplishment in "inventing" the light bulb
was to FIND a material that could be heated so hot that it would glow
brightly and still NOT burn.

He tried hundreds of materials. The one he finally settled on, and the
one still used in incandescent bulbs today, is Tungsten ... Element #74,
a metal that doesn't melt until it's heated to almost 6,200 degrees !

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Explain what is the meaning of positive connotation and negative connotation.

uysha [10]

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4 years ago

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How does friction affect the distance of an object?

stiv31 [10]

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3 years ago

Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i

Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0

3 years ago

An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte

telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

$I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2$

a. The angular acceleration is Torque divided by moments of inertia:

$\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2$

b. 5 revolution would be equals to $10\pi$ rad, or 31.4 rad. Since the engine just got started

$\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5$

$\omega = \sqrt{1893.5} = 43.5 rad/s$

c. Work done during the first 5 revolution would be torque times angular displacement:

$W = T*\theta = 1930 * 31.4 = 60633 J$

d. The time it takes to spin the first 5 revolutions is

$t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s$

The average power output is work per unit time

$P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W$ or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

$P_i = T*\omega = 1930*43.5=83983 W$ or 84 kW

7 0

3 years ago

Suppose a moving car has 2000 J of kinetic energy. If the carʹs speed doubles, how much kinetic energy would it then have?

Bond [772]

Answer:

option A

Explanation:

given,

Kinetic energy of the car = 2000 J

speed of the car is doubled

we know,

$KE_1 = \dfrac{1}{2}mv^2$

$2000= \dfrac{1}{2}mv^2$ ........(1)

now, speed of the car is doubled

v' = 2 v

$KE_2 = \dfrac{1}{2}mv'^2$

$KE_2 = \dfrac{1}{2}m(2v)^2$

from equation (1)

$KE_2 = 4\times \dfrac{1}{2}m(v)^2$

$KE_2 = 4\times 2000$

$KE_2 = 8000\ J$

Hence, the Kinetic energy would be equal to 8000 J.

The correct answer is option A.

8 0

3 years ago