Answer:
377 m
Explanation:
number of turns, N = 65
θ = 36°
B1 = 200 micro Tesla
B2 = 600 micro tesla
t = 0.4 s
induced emf, e = 80 mV
Let a be the side of the square coil.
a = 1.45 m
Total length of the wire, L = N x 4a = 65 x 4 x 1.45 = 377 m
Thus, the length of the wire is 377 m.
Answer:
No, the distance from the last stop to the school and the time it takes to travel that distance are required.
C is correct because they would repel each other A is wrong be they wouldn't repel And B is wrong because they shouldn't be repelling each other
The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>
Solution:
As we know that displacement is calculated in centimeters and the unit of time is second.
The average velocity for the first interval [1,2] is given
Δs / Δt = s (t2) - s (t) / t2 - t1
Δs / Δt = 2sin2 π + 3cos 2 π - ( 2sin π + 3cos π ) / 2 - 1
Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1
Δs / Δt = 6 cm/s
Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s
If you need to learn more about displacement click here:
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The complete question is:
The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1
Answer:
41.4* 10^4 N.m^2/C
Explanation:
given:
E= 4.6 * 10^4 N/C
electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field
then electric flux = ∫ E*n dA
= ∫ 4.6 * 10^4 * 3*3
= 41.4* 10^4 N.m^2/C
