If Calcium lost two electrons, it would have the same number of electrons as Argon which has 18 electrons.
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Answer:</h2>
(a) 3.96 x 10⁵C
(b) 4.752 x 10⁶ J
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Explanation:</h2>
(a) The given charge (Q) is 110 A·h (ampere hour)
Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;
=> Q = 110A·h
=> Q = 110 x 1A x 1h [1 hour = 3600 seconds]
=> Q = 110 x A x 3600s
=> Q = 396000A·s
=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C
Therefore, the number of coulombs of charge is 3.96 x 10⁵C
(b) The energy (E) involved in the process is given by;
E = Q x V -----------------(i)
Where;
Q = magnitude of the charge = 3.96 x 10⁵C
V = electric potential = 12V
Substitute these values into equation (i) as follows;
E = 3.96 x 10⁵ x 12
E = 47.52 x 10⁵ J
E = 4.752 x 10⁶ J
Therefore, the amount of energy involved is 4.752 x 10⁶ J
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
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
