Answer:
the voltage drop across this same diode will be 760 mV
Explanation:
Given that:
Temperature T = 300°K
current = 100 μA
current = 1 mA
forward voltage = 700 mV = 0.7 V
To objective is to find the voltage drop across this same diode if the bias current is increased to 1mA.
Using the formula:
where;
= 0.7
Suppose n = 1
Then;
760 mV
Thus, the voltage drop across this same diode will be 760 mV
Answer:
The answer is below
Explanation:
i) Since the length of the second clock (radius) is 14 cm = 0.14 m, the distance covered by the second hand in one revelution is:
Distance covered = 2πr = 2π(0.14) = 0.88 m
The time taking to complete one revolution = 60 seconds, hence;
Speed = distance covered in one revolution / time take o complete a revolution
Speed = 0.88 m / 60 s = 0.0147 m/s
ii) Distance covered in 150 s = speed * 150 s = 0.0147 * 150 = 2.2 m
iii) Displacement in 150 seconds = distance from initial position to final position
At 150 s, the hand has covered 2 revolutions and moved 30 s. Hence:
Displacement in 150 seconds = speed * 30 s = 0.0147 * 30 = 0.44 m
Can I send you the link ?
D it’s definitely D if I’m wrong sorry…
Answer:
63N, 45N
Explanation:
Given:
Tension 1 = T1
We have the following free body equations:
F - T1 = 100*0.3 -(I)
T1 - T2 = 60*0.3 -(II)
T2 = 150*0.3 -(III)
Adding the three above equations, we have
F - T1 + T1 - T2 + T2 = (100*0.3) + (60*0.3) + (150*0.3)
Hence, F = 93N
Hence, substituting into equation I we have
T1 = Force by bar on lighter cart = 93 - (100*0.3) = 63N
Hence, force that heavier cart exerts on bar
T2 = 150*0.3 = 45N