A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the a
1 answer:
Answer:
C = 4,174 10³ V / m^{3/4}, E = 7.19 10² / ∛x, E = 1.5 10³ N/C
Explanation:
In this problem, we are tasked with determining the constant value and the generated electric field.
We will begin with computing the constant C:
V = C
C = V / x^{4/3}
C = 220 / (11 10⁻²)^{4/3}
C = 4,174 10³ V / m^{3/4}
Next, we will find the electric field by utilizing the formula:
V = E dx
E = dx / V
E = ∫ dx / C x^{4/3}
E = 1 / C x^{-1/3} / (- 1/3)
E = 1 / C (-3 / x^{1/3})
We consider the evaluation from the lower limit x = 0 where E = E₀ = 0 to the upper limit x = x, resulting in E = E:
E = 3 / C (0- (-1 / x^{1/3}))
E = 3 / 4,174 10³ (1 / x^{1/3})
E = 7.19 10² / ∛x
Substituting x = 0.110 cm:
E = 7.19 10² /∛0.11
E = 1.5 10³ N/C
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Explanation:
The measurement of pressure is indicated as where p denotes the pressure,
signifies density, and h represents height
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, and height =1.163 m
Answer:
2.87 m
Explanation:
Given parameters:
Mass of the ball (m) = 60 g = 0.06 kg
Height of the tube (h) = 0.70 m
Force applied on the ball by compressed air (F) = 3.0 N
Initial velocity of the ball (u) = 0 m/s (Assumed)
Final velocity of the ball at the tube's exit (v) =?
Acceleration of the ball (a) =?
The ball's weight is derived from multiplying mass and gravity. Therefore,
Weight (W) =
Thus, the total force acting on the ball equals the net of upward force minus the weight.
Net force = Air force - Weight
According to Newton's second law, net force equals the mass multiplied by acceleration.
Acceleration (a) is calculated as 40.2 m/s².
Using the motion equation, we find:
Let’s denote the maximum height achieved as 'H'.
Next, we apply the principle of energy conservation from the pipe's peak to the maximum height.
A decrease in kinetic energy equals an increase in potential energy.
Substituting the values, we solve for 'H', yielding:
Hence, the ball ascends to a height of 2.87 m above the top of the tube.