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4 years ago

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Answer the question below. Use the rubric in the materials for help if needed. Use details to support your response.

1 answer:

butalik [34]4 years ago

5 0

Answer:

umm soo no wait what

Explanation:

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Cyclist A is moving at 20.0 m/s whereas cyclist B is moving at 12.0 m/s in the same direction and is initially ahead ofA. When t

stealth61 [152]

Answer:

$v_{fA} = 28 \frac{m}{s}$

Explanation:

The kinematic parameters from the moment the two cyclists begin to accelerate until they meet are:

Initial parameters:

$v_{oA} = 20 m/s$ : Initial speed of cyclist A

$v_{oB} = 12 m/s$ : Initial speed of cyclist B

Final parameters:

$v_{fB} = 20 m/s$ : Final speed of cyclist B

$d_{A} = d_{B}$

distance of cyclist A = distance of cyclist B

$t_{A} =t_{B} = 12s$

time of cyclist A = time of cyclist B

Cyclist B Kinematics

$v_{fB} = v_{oB} +a_{B} *t\\$

$36= 12 + a_{B} *12$

$a_{B} = \frac{36-12}{12}$

$a_{B} = 2 \frac{m}{s^{2} }$

$d_{B} =( v_{oB})*( t)+( \frac{1}{2} )*(a_{B})* (t)^{2}$

$d_{B} =( 12*(12)+( \frac{1}{2} )*(2)* (144)}$

$d_{B} = 288m$

Cyclist A Kinematics

$d_{A} =( v_{oA})*( t)+( \frac{1}{2} )*(a_{A})* (t)^{2}$

$d_{A} =(20*( 12)+( \frac{1}{2} )*(a_{A})* 144}$

$d_{A} = 240 + 72*(a_{A})$

$288=240+72*(a_{A} )$

$a_{A} = \frac{288-240}{72}$

$a_{A} = 0.67 \frac{m}{s^{2} }$

$v_{fA} = v_{oA} + a_{A} * t$

$v_{fA} = 20 + 0.67*12$

$v_{fA} = 28 \frac{m}{s}$

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Answer:

entropy

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masya89 [10]

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What does mean kinematics

Travka [436]

Answer:

the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion.

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Which best describes how magnesium and oxygen bond?

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They form an ionic bond by exchanging two electrons.

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