Answer:
An acute injury is sudden and severe such as a broken bone. A chronic injury develops and worsens over an extended period of time like shin splints
Explanation:
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
Answer:
The electric field at origin is 3600 N/C
Solution:
As per the question:
Charge density of rod 1, 
Charge density of rod 2, 
Now,
To calculate the electric field at origin:
We know that the electric field due to a long rod is given by:
Also,
(1)
where
K = electrostatic constant = 
R = Distance
= linear charge density
Now,
In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.
At x = - 1 cm = - 0.01 m:
Using eqn (1):
(towards)
Now, at x = 1 cm = 0.01 m :
Using eqn (1):
(towards)
Now, the total field at the origin is the sum of both the fields:
Weight of the child m = 50 kg
Radius of the merry -go-around r = 1.50 m
Angular speed w = 3.00 rad/s
(a)Child's centripetal acceleration will be a = w^2 x r = 3^2 x 1.50 => a = 9 x
1.5
Centripetal Acceleration a = 13.5m/sec^2
(b)The minimum force between her feet and the floor in circular path
Circular Path length C = 2 x 3.14 x 1.50 => c = 3 x 3.14 => C = 9.424
Time taken t = 2 x 3.14 / w => t = 6.28 / 3 => t = 2.09
Calculating velocity v = distance / time = 9.424 / 2.09 m/s => v = 4.5 m/s
Calculating force, from equation F x r = mv^2 => F = mv^2 / r => 50 x (4.5)^2
/ 1.5
F = 50 x 3 x 4.5 => F = 150 x 4.5 => F = 675 N
(c)Minimum coefficient of static friction u
F = u x m x g => u = F / m x g => u = 675/ 50 x 9.81 => 1.376
u = 1.376
Hence with the force and the friction coefficient she is likely to stay on merry-go-around.
