KCL stands for Potassium Chloride
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yeah i did, hope you understand
Answer:
0.88
Explanation:
velocity (v) = 22 m/s
radius (r) = 56 m
acceleration due to gravity (g) = 9.8 m/s^{2}
What minimum coefficient of static friction between the tires and road is necessary for the car to round the curve without slipping?
for the car to successfully go round the curve, the frictional force must be equal to or greater than the normal force, therefore
Normal force = frictional force
ma = μmg
where
- μ = coefficient of friction
- g = acceleration due to gravity
- a = = = 8.6
- putting in all required values the equation now becomes
8.6m = μm x 9.8
8.6 = 9.8 x μ
μ = 8.6/9.8 = 0.88
The work done by the two forces in moving the chandelier to the given distance is 5,091.2 J.
<h3>
Work done by a single force on the chandelier</h3>
The work done by a single force on the chandelier is determined by taking the net horizontal force applied on the chandelier over the given distance.
where;
is frictional force = 0 (smooth surface).
<h3>Work done by the two forces</h3>
When the two forces combine to pull the chandelier, the total work done will be shared by the two forces.
Thus, the work done by the two forces in moving the chandelier to the given distance is 5,091.2 J.
Learn more about work done here: brainly.com/question/8119756
Answer:
2.7 x 10^-14 m
Explanation:
E =8.8 MeV = 8.8 x 1.6 x 10^-13 J
q = 2 e = 2 x 1.6 x 10^-19 C
Q = 82 e = 82 x 1.6 x 10^-19 C
Let d be the distance of closest approach
E = k Q q / d
Where, K = 9 x 10^9 Nm^2 / C^2
d = k Q q / E
d = (9 x 10^9 x 82 x 1.6 x 10^-19 x 2 x 1.6 x 10^-19) / (8.8 x 1.6 x 0^-13)
d = 2.7 x 10^-14 m