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never [62]
3 years ago
8

How is a manipulated variable different from a controlled variable?

Physics
1 answer:
exis [7]3 years ago
8 0

Answer its d

Hope this helps

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If our solar system's Sun was half the mass as it is now, what would happen to the orbit of the Earth
avanturin [10]
Essentially, an increase in mass means an increase in gravity, while a decrease in mass means a decrease in gravity. So if the Sun spontaneously lost half its mass (gasp, shudder, the horror!) then its gravitational pull on Earth would lessen as well, ultimately changing the very nature of our orbit around the Sun.
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Please Mark Brainliest!
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3 years ago
Halp me please i will give brainly
Delicious77 [7]

Answer: It should be the 3rd option down!

Explanation:

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You would have the largest mass of gold if your chunk of gold weighed 1 N on
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The answer is 1 pound

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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
6. A wave has a frequency of 600 Hz and is traveling at 300 m/s. What is its<br> wavelength?
Luda [366]

Answer:

0.5m

Explanation:

v=f×lamda

v is 300m/s, f is 600Hz, lamda is ?

lamda=v/f

lamda=300/600

lamda =3/6=1/2m

5 0
3 years ago
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