Answer:

Explanation:
Given that,
Initial angular velocity, 
Acceleration of the wheel, 
Rotation, 
Let t is the time. Using second equation of kinematics can be calculated using time.

Let
is the final angular velocity and a is the radial component of acceleration.

Radial component of acceleration,

So, the required acceleration on the edge of the wheel is
.
Answer:
5.25 m
Explanation:
Given;
The height equation h;
h=-x^2+3x+3
Where;
h = the height above water
x = horizontal distance from the end of the board
The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.
differentiating the equation h.
dh/dx = h' = -2x + 3 = 0
Solving for x;
2x = 3
x = 3/2
Substituting into the function h;
h max = -x^2+3x+3
h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =
h max = 21/4 = 5.25 m
Answer:

Explanation:
<u>Given:</u>
- Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
- Distance of separation between the plates, d = 1.0 cm = 0.01 m.
- Minimum value of electric field that produces spark,

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

where,
= surface charge density of the plate of the capacitor =
.
= magnitude of the charge on each of the plate.
= surface area of each of the plate =
= electrical permittivity of free space, having value = 
For the minimum value of electric field that produces spark,

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.
A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
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