Answer:
Average speed is 60 km/hour
Explanation:
When we need to calculate average speed, we use this equation:

Where:
position at the beginning
at the end


Then: 

Finally V = 60 km/hour
Answer:
A) the maximum acceleration the boulder can have and still get out of the quarry
B) how long does it take to be lifted out at maximum acceleration if it started from rest
Explanation:
A)
let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.
the weight of the chain is:
and maximum tension is 
total mass and weight is :


∑



B)
maximum acceleration

using 
to solve for t


Answer:
D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force
Explanation:
For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.
Answer:
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