Answer:
66.375 x 10⁻⁶ C/m
Explanation:
Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as
;
∅ = Q / ε₀ -----------------(i)
Where;
∅ = 7.5 x 10⁵ Nm²/C
ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²
Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;
7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)
Solve for Q;
Q = 7.5 x 10⁵ x 8.85 x 10⁻¹²
Q = 66.375 x 10⁻⁷ C
Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e
L = Q / l ----------------------(ii)
Where;
Q = 66.375 x 10⁻⁷ C
l = length of the rod = 10.0cm = 0.1m
Substitute these values into equation (ii) as follows;
L = 66.375 x 10⁻⁷C / 0.1m
L = 66.375 x 10⁻⁶ C/m
Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.
Answer:
a
b
The value is 
Explanation:
From the question we are told that
The mass is 
The spring constant is 
The instantaneous speed is 
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So
Here a is the amplitude of the subsequent oscillations
=> 
=> 
=>
Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point
=> 
=>
They are blue because of hydrogen helium and methane
Answer:
4 m/s
Explanation:
m1 = m2 = m
u1 = 20 m/s, u2 = - 12 m/s
Let the speed of composite body is v after the collision.
Use the conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
m x 20 - m x 12 = (m + m) x v
20 - 12 = 2 v
8 = 2 v
v = 4 m/s
Thus, the speed of teh composite body is 4 m/s.
