Answer:
From certain assumptions that the walking speed is 2 m/s, and the stop time is 0.1 s the acceleration would be -20 m/s
Explanation:
Using the average acceleration formula:
where
and
are the changes in the speed and time respectively.
We have by assuming that the walking speed is 2 m/s and the stop time is 0.1s which is equal to the change in time during the stopping.
, where
are the initial speed and final speed respectively, and 
Plugging the previous in the average acceleration formula we get
where the minus sign indicates an acceleration in the opposite direction of the motion (or in other word opposite to the speed's direction).
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A man pushes a couch across the room is the answer!
Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:

Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution
