Answer:
Explanation:
Let m be mass of each sphere and θ be angle, string makes with vertex in equilibrium.
Let T be tension in the hanging string
T cosθ = mg ( for balancing in vertical direction )
for balancing in horizontal direction
Tsinθ = F ( F is force of repulsion between two charges sphere)
Dividing the two equations
Tanθ = F / mg
tan17 = F / (7.1 x 10⁻³ x 9.8)
F = 21.27 x 10⁻³ N
if q be charge on each sphere , force of repulsion between the two
F = k q x q / r² ( r is distance between two sphere , r = 2 x .7 x sin17 = .41 m )
21.27 x 10⁻³ = (9 X 10⁹ x q²) / .41²
q² = .3973 x 10⁻¹²
q = .63 x 10⁻⁶ C
no of electrons required = q / charge on a single electron
= .63 x 10⁻⁶ / 1.6 x 10⁻¹⁹
= .39375 x 10¹³
3.9375 x 10¹² .
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.11
Explanation:
Given;
initial speed, u = 9.3 m/s
sliding distance, S = 42 m
From equation of motion we determine the acceleration;
v² = u² + 2as
0 = (9.3)² + (2x42)a
- 84a = 86.49
a = -86.49/84
|a| = 1.0296
= ma
where;
Fk is the frictional force
μk is the coefficient of kinetic friction
N is the normal reaction = mg
μkmg = ma
μkg = a
μk = a/g
where;
g is the gravitational constant = 9.8 m/s²
μk = a/g
μk = 1.0296/9.8
μk = 0.11
Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11
5 seconds is a poor time to ask about, because the speed abruptly changes at exactly 5 seconds.
Up until that time, the speed has been 1 m/s. And then, at exactly 5 seconds, it becomes zero.
_________
It's also a poor question because speed is calculated from the distance covered, but the graph shows displacement, not distance. You can't really tell the distance covered from a displacement graph.
For example, if an object happens to be moving in a circle around the place where it started, then the total distance covered keeps increasing, but its displacement is constant.
Answer A sounds about right