The force of friction is given by:
f = μR, where μ is the friction coefficient and R is the reaction force, which will be equal to the weight.
100 = μ x 130
μ = 0.77
data which is expressed in form of following way

here in above expression
= true value
= uncertainty in the value
now the relative uncertainty is given as

now by above formula we can say
a) 2.70 ± 0.05cm
here
True value = 2.70
uncertainty = 0.05
Relative uncertainty =
= 0.0185
b) 12.02 ± 0.08cm
here
True value = 12.02
uncertainty = 0.08
Relative uncertainty =
= 0.00665
Answer:
The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
Explanation:
P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.
Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.
Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
Answer:
Sorry don't know the answer
Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;

Where;
d is the separation or distance between the two parallel plates;

Therefore, the necessary separation between the two parallel plates is 0.104 mm