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pogonyaev
3 years ago
13

What type of material is good at transferring heat?

Physics
1 answer:
MatroZZZ [7]3 years ago
5 0
The answer is A

Materials that are good conductors of thermal energy are called thermal conductors. Metals are very good thermal conductors. Materials that are poor conductors of thermal energy are called thermal insulators. Gases such as air and materials such as plastic and wood are thermal insulators
You might be interested in
NGSS Physics
vovangra [49]
Answer: 3.5 seconds

EXPLANATION:

Using the formula:
v = u + at
And taking the upwards direction as positive, we have the following information:

u = 35 m/s
a = -10m/s^2 (this is acceleration due to gravity)

At the top of its path, the apple will have a velocity of 0 m/s, therefore:

v = 0m/s

Once you substitute everything into the formula, you get:

0 = 35 + (-10)t

Therefore, t = 35/10 = 3.5 seconds
6 0
3 years ago
When does sea breeze occur
Vladimir79 [104]
<span>Sea breeze can happen during hot summer days because of the uneven  heating rates of water and land. The land surface heats up faster than the surface of the water during the day.  At this rate, the air above the land grows warmer than the air atop the ocean. Warmer air is always lighter than cooler air. As a consequence, warm air is pushed upward causing it to rise. With this, warmer air rises over the land. As warm air rises over the land, cooler air over the ocean flows over the land surface to change or replace the rising warm air.</span>


3 0
3 years ago
Read 2 more answers
If a star with an absolute magnitude of -5 has an apparent magnitude of +5 ,then its distance is
klio [65]
You asked a question.  I'm about to answer it. 
Sadly, I can almost guarantee that you won't understand the solution. 
This realization grieves me, but there is little I can do to change it. 
My explanation will be the best of which I'm capable.


Here are the Physics facts I'll use in the solution:

-- "Apparent magnitude" means how bright the star appears to us.

-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).

-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by  ⁵√100 .
That's about  2.512... .  

-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .

-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).

That's all the Physics.  The rest of the solution is just arithmetic.
____________________________________________________

-- The star in the question would appear M(-5) at a distance of
32.6 light years. 

-- It actually appears as a M(+5).  That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.

-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .

-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is

(32.6) · (100)^(1) light years

= (32.6) · (100) light years

=  approx.  3,260 light years .   (roughly 1,000 parsecs)


I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
6 0
2 years ago
I REALLY NEED HELP WITH PHYSICS ASAP!!!<br>Vf^2 = v0^2 + 2a (xf - x0)<br><br><br>Solve for a
ELEN [110]

Answer:

a. solve for a

vf ^{2} = vo ^{2}  + 2a(xf - xo) \\ 2a(xf - xo) = vf^{2}    -  vo ^{2}   \\ a =  \frac{vf^{2} - vo^{2}  }{2(xf - xo)}    \\ a =  \frac{vf ^{2} - vo ^{2}  }{2xf - 2xo}

I hope I helped you ^_^

7 0
3 years ago
Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als
victus00 [196]

Electric field lines always begin at positive charges (or at infinity) and end at negative charges (or at infinity).

One could also say that the lines we use to represent an electric field indicate the direction in which a positive test charge would initially move when released from rest.

6 0
3 years ago
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