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Daniellelovee
 one year ago
Suppose you are using a system of units based on the centimeter (cm) for length, the gram (g) for mass, and the second (s) for time. The kinetic energy that an object has because of its motion is equal to half its mass times its speed squared, or KE=1/2mv2. In this system of units, what units would be used to express KE?
g*cm2/s2
g*cm/s
kg*cm2/s2
g*cm3/s
Daniellelovee
 one year ago
Suppose you are using a system of units based on the centimeter (cm) for length, the gram (g) for mass, and the second (s) for time. The kinetic energy that an object has because of its motion is equal to half its mass times its speed squared, or KE=1/2mv2. In this system of units, what units would be used to express KE? g*cm2/s2 g*cm/s kg*cm2/s2 g*cm3/s

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0m is mass v is velocity Mass would have to be in grams according to your question Cm would have to be your distance or length Time would have to be in seconds But velocity is distance (cm)/time (s) so units for velocity is cm/s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Show me your work. Dont tell me.

Daniellelovee
 one year ago
Best ResponseYou've already chosen the best response.0ok so half of grams time speed squared would be g/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Stop mentioning answers until you solved it please.

Daniellelovee
 one year ago
Best ResponseYou've already chosen the best response.0ok but Im sure is A now because it seems similar to what the problem is saying

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0grams times speed is \[g \times \frac{ m }{ s }\]

Daniellelovee
 one year ago
Best ResponseYou've already chosen the best response.0yes but the speed on the problem is squared

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you solve it algebraically, I can confirm your answer. Tell me which part you are lost in and I will help.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yep that is logical, but it might not always be correct.

Daniellelovee
 one year ago
Best ResponseYou've already chosen the best response.0nvm Im not lost anymore, I got this, and thank you for your help :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No problem! Good luck on the rest.
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