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Masteriza [31]
3 years ago
11

Which state of matter is best at conducting heat?

Physics
2 answers:
Black_prince [1.1K]3 years ago
7 0
<span>A state f matter best at conducting heat would be solids.</span>
jek_recluse [69]3 years ago
4 0
Solids are the best at conducting heat. 
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Where are you likely to find a cooling plant located in the ductwork? A. A medium-sized office B. A large factory C. A single fa
Rufina [12.5K]

Answer:

i think it would be B, a large factory

Explanation:

8 0
2 years ago
The magnitude of the tidal force between the International Space Station (ISS) and a nearby astronaut on a spacewalk is approxim
vovikov84 [41]

Answer:

F = 4.47 10⁻⁶ N

Explanation:

The expression they give for the strength of the tide is

      F = 2 G m M a / r³

Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg

They ask us to perform the calculation

      F = 2 6.67 10⁻¹¹ 135  5.98 10²⁴ 13 / (6.79 10⁶)³

      F = 4.47 10⁻⁶ N

This force is directed in the single line at the astronaut's mass centers and the space station

4 0
3 years ago
How do observing and inferring differ?
Illusion [34]
Observing is commenting on what you see e.g: It is raining.
Inferring is drawing conclusions from observations for example you see the road is wet so you might infer that it rained last night.
6 0
2 years ago
Read 2 more answers
Which is a gas at room temperature?
Sedaia [141]
<span>Which is a gas at room temperature?
</span>B) nitrogen

7 0
2 years ago
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A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat
Sever21 [200]

Answer:

Approximately 325 (rounded down,) assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, {\rm J}):

\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}.

Height of the weight (should be in meters, {\rm m}):

\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}.

Energy required to lift the weight by \Delta h = 0.2\; {\rm m} without acceleration:

\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}.

At an efficiency of 0.25, the actual amount of energy required to raise this weight to that height would be:

\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}.

Divide 2.551 \times 10^{5}\; {\rm J} by 784\; {\rm J} to find the number of times this weight could be lifted up within that energy budget:

\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}.

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0
1 year ago
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