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2 years ago

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What is the acceleration of a 32-kg object if a 6.75-N force is applied to it?

1 answer:

professor190 [17]2 years ago

5 0

F=MA
*newton 2nd law*
F=6.75N
M=32kg

A= F/M
A=6.75/32
equal
a.0.21m/s2

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Explanation:

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3 years ago

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3 years ago

Read 2 more answers

If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is

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Answer:

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Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec

The object will return with the same speed that it had initially so the object

starts out with a downward speed of 44 ft/sec

Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2

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3 years ago

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saveliy_v [14]

Answer:

The speed it reaches the bottom is

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$w=27.74rad/s$

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