If the maximum emf of the ac generator is 20 V and the maximum potential difference across the resistor is 16 V Then the maximum potential difference across the inductor is 4 V.
Calculation:
Step-1:
It is given that the RL circuit is connected to a 20 V ac generator. The maximum potential difference across the resistor is 16 V. It is required to find the maximum potential drop across the inductor.
Step-2:
The maximum emf of the generator is equal to the sum of the maximum potential difference across the resistor and the maximum potential difference across the inductor.
Therefore,
The maximum potential difference across the inductor + Maximum maximum potential difference across the resistor = Maximum emf of the generator
Thus,
Maximum maximum potential difference across the inductor + 16 V = 20 V
Therefore,
Maximum maximum potential difference across the inductor = 20 V - 16 V = 4 V
Learn more about potential differences across resistor and inductor here,
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B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string = 
linear density of the string = 
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m
question 2 answer is ALL OF THE ABOVE
question 3 answer is WARM AND MOIST.
a. Sweet corn and possibly d. okra.
