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bagirrra123 [75]
3 years ago
5

Must new substance be formed when you observe property

Physics
1 answer:
d1i1m1o1n [39]3 years ago
6 0
Of course not !
You can observe the color of a rose, measure the length of
your house, take the temperature of your hot chocolate, and
measure your own weight, without creating any new substances.
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Without fluid friction, all objects accelerate at?
lys-0071 [83]

Answer:

Acceleration of gravity=9.8m/s/s

Explanation:

Newton's Second Law-acceleration is proportional to the net force acting on an object.

All objects usually free fall at the same acceleration of 9.8m/s/s-this regardless of their mass. This acceleration is known as acceleration of gravity.

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Question is attatched!
lina2011 [118]
All it does is lets him pull in a more convenient direction to raise the load. It has no effect on the required force.
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MATCH THESE ^-^ Match Newton's law with the correct statement.
ikadub [295]
1 and A
2 and B
3 and D
4 and C

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What might be a reason conflict arises?
Katena32 [7]

D- all of the above

Explanation:

Conflict is a clash between different ideas and people.

  • It can be as a result of the misunderstanding of another person's point of view. This is one of the leading causes of conflict. A party can take another party's view verbatim without understanding the context behind a reasoning. Through this, conflict can ensue.
  • It can also be due to non-compatible personalities. This can be between an extrovert and introvert or a blend of them.
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Learn more;

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
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