A chair of weight 90.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F =
38.0 N directed at an angle of 42.0 degrees below the horizontal and the chair slides along the floor.
Using newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Using newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
1 answer:
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Answer:
n=6.56×10¹⁵Hz
Explanation:
Given Data
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
To find
Revolutions per second
Solution
Let F be the force of attraction
let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by
F=mω²r......................where ω=2π n
F=m4π²n²r...............eq(i)
as the values given where
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
we have to find n from eq(i)
n²=F/(m4π²r)
Answer:
24.3 m/s
Explanation:
1 kmh = 0.27 m/s, that makes a conversion ratio of 0.27/1kmh
x
The "kmh" n the top and bottom cancel out. And then you just multiply the top 90 x 0.27 and the bottom 1 x 1 to get
and since its over 1 its just 24.3 m/s