The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.
To find the answer, we have to know more about the ideal diode.
<h3>
What is an ideal diode?</h3>
- A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
- A semiconductor diode is the kind of diode that is used the most commonly.
- It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.
<h3>How to find the current in ideal diode?</h3>
- Here we have given with the values,
- We have the expression for current in mA of the ideal diode with forward biased voltage drop as,
Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.
Learn more about the ideal diode here:
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A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
Answer:
Option D.
Explanation:
The correct answer is Option D.
The shrinkage of the bridge material is because of thermal contraction.
Thermal contraction and thermal expansion are the phenomena of the bridge material which takes place due to the change in temperature of the atmosphere.
When the temperature of the surrounding increases expansion of the bridge material takes place and when temperature decreases the contraction of the material takes place.
This phenomenon sometimes damages the structure because due to continuous expansion and contraction of materials strength of the bridge decreases.
Answer:
You were a freeloader of my questions, so I'll be one too.
Answer:
(a) W=1.20×10⁴J
(b) U= -5.46×10⁴J
(c) Q= -4.26×10⁴J
Explanation:
Given that student does 1.20×10⁴J work
(a) W=1.20×10⁴J
Work done by student,so positive sign
During the process, his internal energy decreases by 5.46×10⁴J.
(b) U= -5.46×10⁴J.
As the Energy decreases therefore negative sign
For (c) Q
We know the formula