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UkoKoshka [18]
3 years ago
14

HW 3 a : Power Supply LTspice Design a resistor/LED circuit such that the diode will be brightly let when V out i s 5 volts. Ass

ume the diode has a forward voltage of 3.65V and a forward current of 20mA, when the diode is brightly emitting light. Note: You need to round your resistors to a whole number, as you cannot purchase non - whole numbers resistors.
Physics
1 answer:
nadya68 [22]3 years ago
4 0

Answer:

68 ohm

Explanation:

Output Voltage, Vout = 5 V

Forward voltage, Vf = 3.65 V

Forward current, If = 20 mA = 0.02 A

Let the resistance is R.

Use the formula

I_{f}=\frac{V_{out}-V_{f}}{R}

0.02=\frac{5-3.65}{R}

R = 67.5 ohm

R = 68 ohm ( approx.)

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43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a
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Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N

The frictional force that acts on the sliding bear is 378.47166 N

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Answer:

(c)  As 'd' becomes doubled, energy decreases by the factor of 2

Explanation:

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As capacitor remains connected to the battery so V remains constant. As can be seen from (1) that energy is inversely proportional to the separation between the plates so as 'd' becomes doubled, energy decreases by the factor of 2.

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