The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
Answer:
Explanation:
If a number of less than 1, then the number has a decimal point like
0.085, 0.008 e.t.c.
The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.
But if there a zero between the none zero e.g. 0.0087056
Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6
But if the zero is not in between the none zero digit, then the zero is insignificant
E.g 0.05800
The last two zero is insignificant, the significant number is 5 and 8
So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.
<span>Amplitude is the correct answer. I hope this helps.</span>
Answer:
A.) 27000 kgm/s
18000 kgm/s
B.) Va = 22 m/s
C.) 19800 kgm/s
25200 kgm/s
Explanation: Given that the velocity of A and B are 30 m/s and 20 m/s. And of the same mass M = 9 × 10^5g
M = 9×10^5/1000 = 900 kg
A.) Initial momentum of A
Mu = 900 × 30 = 27000 kgm/s
Initial momentum of B
Mu = 900 × 20 = 18000 kgm/s
B.) if they have an accident and then the velocity of the B is 28 m/s, find out velocity of A.
Momentum before impact = momentum after impact
Given that Vb = 28 m/s
27000 + 18000 = 900Va + 900 × 28
45000 = 900Va + 25200
900Va = 45000 - 25200
900Va = 19800
Va = 19800/900
Va = 22 m/s
C.) Momentum of A after impact
MV = 900 × 22 = 19800 kgm/s
Momentum of B after impact
MV = 900 × 28 = 25200 kgm/s
