Correct question:
An elevator in a mine shaft is stopped at 25 feet below ground level. And then travels 28 feet downward. What equation can be written to find the final position in feet of the elevator in relations to ground level?
Answer:
-53 feets
Step-by-step explanation:
Initial stop point = 25 feets below ground level
Further distance traveled = 28 feets below ground level
Final position in feet = Initial stop point + further distance traveled
Note : distances or point traveled below ground level or downward is represented as negative.
Final position in feets = (-25)feets + (-28)feets
Final position in feets = - 53 feets
It’s it four I think hope it helps
One tenth is 0.1. Therefore, the answer has to be 2.x. Since 4 is less than 5, the 2 remains the same. The answer is 2.2.
Answer:
18.67 m/s²
Step-by-step explanation:
a = ∆v/ t
a= v - u/t
a = 2.8 - 0 / 0.15
a = 2.8 / 0.15
a = 18.67 m/s²
