A 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal forc
1 answer:
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Answer:
0 N/C
Explanation:
Parameters given:
The distance between and
is
Electric field is given as
r = 15/2 = 7.5cm = 0.075m
The electric field at their midpoint due to is:
=
The electric field at the midpoint due to is:
=
The net electric field will be:
E = +
E = +
E = 0 N/C
Answer:
(a) The elevator has an acceleration of 2.14 m/s² and direction is downward
(b) The acceleration is a=2.14 m/s²
Explanation:
Let +y be upward direction
For Part (a)
The mass of student (whose weight is 550N) is:
ms=w/g
ms=550/9.8=56.1 kg
Apply ∑Fy=ma
Solve for acceleration a
The elevator has an acceleration of 2.14 m/s² and direction is downward
Part(b)
For the acceleration is the scale reads 670 N
For n=670N
So