Answer:
The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].
Explanation:
Given that,
Area of cube = 48 cm²
Charge = 28.7 nC
We need to calculate the flux of the electric field through the surface
Using formula Gauss's law
The electric flux through any closed surface,
Where, q = charge
Put the value into the formula
Hence, The flux of the electric field through the surface is 3.24\times10^{3}\ Nm^/C[/tex].
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Answer:
The orbital velocity
The period is
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the planet , this is mathematically represented as
=>
Here G is the gravitational constant with value
is the mass of with value
r is the is distance from the center of the to the outer edge of the A ring
i.e r = R + D
Here R is the radius of the planet with value
D is the distance from the equator to the outer edge of the A ring with value
So
=>
So
=>
=>
Generally the orbital velocity is mathematically represented as
=>
=>
Generally the period is mathematically represented as
=>
=>
Answer:
The orbital velocity
The period is
Explanation:
Generally centripetal force acting ring particle is equal to the gravitational force between the ring particle and the , this is mathematically represented as
Answer:
K = 36 J
Explanation:
In this exercise we must use the conservation of mechanical energy at two points at initial x = 0 and the end point of maximum compression of the spring
Initial x = 0
Em₀ = K = ½ m v²
Final point of maximum compression
= Ke = ½ k x²
Em₀ =
K = ½ k x²
We put the kinetic energy because it is a data, let's look for the spring constant
k = 2 K / x²
k = 2 9 / d²
k = 18 / d²
Let's calculate the kinetic energy, so that the compression is x = 2d
K = ½ k x²
K = ½ (18 / d²) (2d)²
K = ½ 18/d² 4d²
K = 36 J