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nasty-shy [4]
3 years ago
7

How many stars are there in an Open Cluster?*

Physics
1 answer:
Step2247 [10]3 years ago
4 0

Answer:

Very few

Most open clusters form with at least 100 stars

Brainly, please.

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B. over the symbol.

Explanation:

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Convert the following statement to the language used by physicists, "I am cold, please turn on the heat."​
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Is a cinder cone volcanoe constructive or deconstructive
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I need homework help
KIM [24]
1) the weight of an object at Earth's surface is given by F=mg, where m is the mass of the object and g=9.81 m/s^2 is the gravitational acceleration at Earth's surface. The book in this problem has a mass of m=2.2 kg, therefore its weight is 
F=mg=(2.2 kg)(9.81 m/s^2)=21.6 N

2) On Mars, the value of the gravitational acceleration is different:g=3.7 m/s^2. The formula to calculate the weight of the object on Mars is still the same, but we have to use this value of g instead of the one on Earth: F=mg=(2.2 kg)(3.7 m/s^2)=8.1 N

3) The weight of the textbook on Venus is F=19.6 N. We already know its mass (m=2.2 kg), therefore by re-arranging the usual equation F=mg, we can find the value of the gravitational acceleration g on Venus: 
g= \frac{F}{m}= \frac{ 19.6 N}{2.2 kg}=8.9 m/s^2

4) The mass of the pair of running shoes is m=0.5 kg. Their weight is F=11.55 N, therefore we can find the value of the gravitational acceleration g on Jupiter by re-arranging the usual equation F=mg: 
g= \frac{F}{m} = \frac{11.55 N}{0.5 kg} =23.1 m/s^2

5) The weight of the pair of shoes of m=0.5 kg on Pluto is F=0.3 N. As in the previous step, we can calculate the strength of the gravity g on Pluto as 
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<span>6) On Earth, the gravity acceleration is </span>g=9.81 m/s^2<span>. The mass of the pair of shoes is m=0.5 kg, therefore their weight on Earth is 
</span>F=mg=(0.5 kg)(9.81 m/s^2)=4.9 N<span>
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5 0
3 years ago
As mass increases what happens to the kinetic energy
AnnyKZ [126]
As mass increases kinetic energy also increases; kinetic energy is directly proportional to mass so whatever is done to either affects the other one the same. i hope this helps :)
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3 years ago
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