Answer:
Explanation:
Energy of system of charges
= k q₁q₂ / r₁₂ + k q₁q₃ / r₁₃ + k q₃q₂ / r₃₂
q₁ , q₂ and q₃ are charges and r₁₂ , r₁₃ , r₃₂ are densities between them
9 x 10⁹ ( 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 + 2x2 x10⁻¹²/ .25 )
= 9 x 10⁹ x 3 x 16 x 10⁻¹²
= 432 x 10⁻³
= .432 J .
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
So,
a) 0 < r < r1 :
We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.
Hence, E = 0 for r < r1
b) r1 < r < r2:
Electric field =?
Let, us consider the Gaussian Surface,
E x 4 =
So,
Rearranging the above equation to get Electric field, we will get:
E =
Multiply and divide by
E = x
Rearranging the above equation, we will get Electric Field for r1 < r < r2:
E= (σ1 x ) /( x )
c) r > r2 :
Electric Field = ?
E x 4 =
Rearranging the above equation for E:
E =
E = +
As we know from above, that:
= (σ1 x ) /( x )
Then, Similarly,
= (σ2 x ) /( x )
So,
E = +
Replacing the above equations to get E:
E = (σ1 x ) /( x ) + (σ2 x ) /( x )
Now, for
d) Under what conditions, E = 0, for r > r2?
For r > r2, E =0 if
σ1 x = - σ2 x
Explanation:
Given
initial velocity(v_0)=1.72 m/s
using
Where v=final velocity (Here v=0)
u=initial velocity(1.72 m/s)
a=acceleration
s=distance traveled
s=0.214 m
(b)time taken to travel 0.214 m
v=u+at
t=0.249 s
(c)Speed of the block at bottom
Here u=0 as it started coming downward
v=1.72 m/s
4200000 is your answer hope this helps
The pressure exerted by a fluid solely relies on the depth or height of the fluid, its density, and the gravitational constant. These three are related in the equation:
Pressure = density x g x height
In the problem, point A is within the block inside the tank. The water above the block is assumed to be 0.6 meters. This gives a point A pressure of:
P = 1000 kg/m^3 * 9.81 m/s^2 * 0.6 m = 5,886 Pa or 5.88KPa