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3 years ago

9

The surface area of a material is a factor that affects heat conductivity. Does heat flow faster through a large surface area or

through a small surface area? Why?
science

1 answer:

Elis [28]3 years ago

4 0

The air flows slower in a bigger space. The air in a small space hit each other heating up, and move faster and faster. is that what your asking?

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A quarter is tossed up from the roof of a skyscraper and hits the sidewalk below. Which of the following graphs best shows the v

laila [671]

Answer:

pic

Explanation:

5 0

2 years ago

Electromagnetic radiation behaves both as particles (called photons) and as waves. Wavelength (λ) and frequency (ν) are related

inessss [21]

<h2>Answer: 136.363 m</h2>

Explanation:

We can find the wavelength of the radiation produced by the microwave oven by using the following given equation:

$c=\lambda.\nu$ (1)

Clearing $\lambda$ :

$\lambda=\frac{c}{\nu}$ (2)

Knowing $\nu=2.20 GHz=2.20(10)^{6}Hz=2.20(10)^{6}s^{-1}$

$\lambda=\frac{3(10)^{8}m/s}{2.20(10)^{6}s^{-1}}$ (3)

$\lambda=136.363m$ This is the wavelength of the radiation produced by the microwave

8 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

3 years ago

In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

Talja [164]

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0

3 years ago

An orchestra is practicing a piece that is to be played at an allegro tempo. The conductor sets a metronome at 160 beats per min

omeli [17]

Answer:

375 ms

Explanation:

the frequency of metronome , f = 160 beats per minute

f = 160 /60 beats per sec

f = 2.67 beats /s

the period of a single beat , T = 1/f

T = 1/2.67 s

T = 0.375 s = 375 ms

the period of a single beat is 375 ms

3 0

3 years ago