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navik [9.2K]
3 years ago
11

Thermal Conductors don't have to be hot to transfer heat, explain a situation when a ice cube would still transfer heat to anoth

er object it is in contact with.
Physics
1 answer:
barxatty [35]3 years ago
6 0
An ice cube would transfer heat to another object whose temperature
is lower than zero°C (32°F).

A block of "dry ice" is sitting there at a temperature of -78°C (-109°F).
An ice cube helps to melt dry ice nice and fast.

If you could find a block of solid nitrogen, its temperature would be
63K (-210°C, -346°F).  An ice cube would transfer heat to that baby
so fast that it would instantly boil.

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Who is this guy and what did he do?
blsea [12.9K]
According to Google, that is Tyler the Creator. He participates in Camp Flog Gnaw which claims to "<span>got a better, more star-studded </span>lineup<span> than most rap festivals can offer." He is also a rapper and appears to be a pretty sick dude.</span>
6 0
3 years ago
A planet has two
lozanna [386]
Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.

<u>Moon #1:</u>  (1.262 days)² / (2.346 x 10^4 km)³

<u>Moon #2:</u>  (orbital period)² / (9.378 x 10^3 km)³

If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.

Cross multiply the proportion:

(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³

Divide each side by (2.346 x 10^4)³:

(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³

               =  0.1017 day²

Orbital period = <u>0.319 Earth day</u> = about 7.6 hours.
7 0
3 years ago
A uniform charge density of 600 nC/m3 is distributed throughout a spherical volume (radius = 14 cm). Consider a cubical (3.2 cm
jolli1 [7]

Answer: The electric flux through this surface is 2.22 Nm^2/C

Explanation: Please see the attachments below

5 0
3 years ago
An LC circuit consists of a 3.14 mH inductor and a 5.08 µF capacitor. (a) Find its impedance at 55.7 Hz. 563.57 Correct: Your an
ANEK [815]

Answer:

a)

z=561.7

b)

z=214.1

Explanation:

L = inductance of the Inductor = 3.14 mH = 0.00314 H

C = capacitance of the capacitor = 5.08 x 10⁻⁶ F

a)

f = frequency = 55.7 Hz

Impedance is given as

z=\frac{1}{2\pi fC} - 2\pi fL

z=\frac{1}{2(3.14) (55.7)(5.08\times 10^{-6})} - 2(3.14) (55.7)(0.00314)

z=561.7

b)

f = frequency = 11000 Hz

Impedance is given as

z= - \frac{1}{2\pi fC} + 2\pi fL

z= - \frac{1}{2(3.14) (11000)(5.08\times 10^{-6})} + 2(3.14) (11000)(0.00314)

z=214.1

4 0
3 years ago
The pressure in a container will stay the same if you ——?
elena55 [62]
I think it’s D...not sure tho !
8 0
3 years ago
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