Average speed = (total distance covered) / (time to cover the distance)
Total distance = (77km + 66km) = 143 kilometers
Time to cover the distance = 2 hours
Average speed = (143 km) / (2 hours) = 71.5 km per hour
Answer:
9.2 amperes
Explanation:
Ohm's law states that the voltage V across a conductor of resistance R is given by 
Here, voltage V is proportional to the current I.
For voltage, unit is volts (V)
For current, unit is amperes (A)
For resistance, unit is Ohms (Ω)
Put R = 12.5 and V = 115 in V=RI

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

