Answer:
The amount of electrons that flow in the given time is 3.0 C.
Explanation:
An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.
i.e I = 
...................(1)
It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.
In the given question, I = 1.5A, t = 2s, find Q.
From equation 1,
Q = I × t
= 1.5 × 2
= 3.0 Coulombs
The amount of electrons that flow in the given time is 3.0 C.
Answer:
a
b
Explanation:
Generally the force of attraction between this two irons is mathematically represented as
![F = \frac{k * [Q_{Li} ] * [Q_{O} ] }{ r^2}](https://tex.z-dn.net/?f=F%20%3D%20%20%5Cfrac%7Bk%20%2A%20%20%5BQ_%7BLi%7D%20%20%5D%20%2A%20%5BQ_%7BO%7D%20%20%5D%20%20%7D%7B%20r%5E2%7D)
Here k is known as the proportionality constant with value 
substituting -2 for 
i.e the charge on oxygen , +1 for 
i.e the charge on Lithium and ![[0.140 + 0.068 ] nm= 0.208 nm = 0.208*10^{-9}](https://tex.z-dn.net/?f=%20%5B0.140%20%2B%200.068%20%5D%20nm%3D%200.208%20nm%20%3D%20%200.208%2A10%5E%7B-9%7D%20%20%20)
for r
So
Generally the force of repulsion will be the magnitude but different direction to the force o attraction
So Force of repulsionn is
Answer:
a) 5.851× 10¹⁰m/s²
b) 2.411×10⁻¹¹s
c) 1.70×10⁻¹¹m
d) 1.661×10⁻²⁷KJ
Explanation:
A proton in the field experience a downward force of magnitude,
F = eE. The force of gravity on the proton will be negligible compared to the electric force
F = eE
a= eE/m
= 1.602×10⁻¹⁹ × 610/1.67×10⁻²⁷
= 5.851× 10¹⁰m/s²
b)
V = u + at
u= 0
v= 1.4106m/s
v= (0)t + at
t= v/a
= 1.4106m/s/5.851 ×10¹⁰
= 2.411×10⁻¹¹s
c)
S = ut + at²
= (o)t + 5.851×10¹⁰×(2.411×10⁻¹¹)²
= 1.70×10⁻¹¹m
d)
Ke = 1/2mv²
= (1.67×10⁻²⁷×)(1.4106)²/2
= 1.661×10⁻²⁷KJ
Answer:
68.8 N
Explanation:
From the question given above, the following data were obtained:
Mass (m) of box = 18 Kg
Coefficient of friction (μ) = 0.39
Force of friction (F) =?
Next, we shall determine the normal force of the box. This is illustrated below:
Mass (m) of object = 18 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal force (N) =?
N = mg
N = 18 × 9.8
N = 176.4 N
Finally, we shall determine the force of friction experienced by the object. This is illustrated below:
Coefficient of friction (μ) = 0.39
Normal force (N) = 176.4 N
Force of friction (F) =?
F = μN
F = 0.39 × 176.4
F = 68.796 ≈ 68.8 N
Thus, the box experience a frictional force of 68.8 N.
