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3 years ago

11

How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead

is Lv = 871 × 103 J/kg.

1 answer:

lara [203]3 years ago

3 0

Answer:

Heat required = mass× latent heat Q = 0.15 × 871 ×

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Explain the difference between the 7 parts of the electromagnetic spectrum.

inn [45]

Answer:

The difference between the two is, well for one

Spectrum: The entire range that the "waves" could be such, as visible light, x-ray's and so on.

Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.

It may confuse you but it makes sense to me (Sorry)

Explanation:

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3 years ago

Do magnets have to touch each other in order to experience a magnetic force

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4 years ago

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In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel

slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=60\times \dfrac{1609.34}{3600}=26.822\ m/s$

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m$

$s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s$

or

$v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s$

The time taken is 2.73414 seconds

The potential energy is given by

$U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J$

The change in potential energy is 467622.66798 J

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3 years ago

A ship's propeller of diameter 3 m makes 10.6 revolutions in 30s. What is the angular velocity of the propeller?

ycow [4]

Answer:

The angular velocity of the propeller is 2.22 rad/s.

Explanation:

The angular velocity (ω) of the propeller is:

$\omega = \frac{\Delta \theta}{\Delta t}$

Where:

θ: is the angular displacement = 10.6 revolutions

t: is the time = 30 s

$\omega = \frac{\Delta \theta}{\Delta t} = \frac{10.6 rev*\frac{2\pi rad}{1 rev}}{30 s} = 2.22 rad/s$

Therefore, the angular velocity of the propeller is 2.22 rad/s.

I hope it helps you!

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3 years ago

You have been hired by a "storm chaser" as an assistant. This individual loves to find locations at which tornadoes and violent

Studentka2010 [4]

Answer:

58.44 C

Explanation:

Electric field is found by

$E=\frac {\sigma}{\epsilon_o}=\frac {Q}{A\epsilon_o}$

Therefore, the charge is

$Q=EA\epsilon_o
$

$Q= 4.00 ✕ 10^{6} N/C*1.65 *10^{6} m^{2}*8.854*10^{-12}= 58.4364 C
$

Therefore, required charge is 58.44 C

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3 years ago