Answer:
1+5x+10x^2+10x^3+5x^4+x^5
Explanation:
You just make it (1+x)(1+x)(1+x)(1+x)(1+x) and multiply it out until it's all one big term.
The electric potential energy of the pair of charges when the second charge is at point b is U at b = +7.3×10⁻⁸J.
Electric potential is the amount of work done in moving point charge from infinity to aspecific point.
Calculation of the electric potential energy:
Here the particle should be moved from the point at the time when the potential energy should be U a
The U b should be the change in the potential energy
W is work done.
(W at a )-(W at b) = U a - U b
U b = U a - (W at a )+(W at b)
So,
= 5.4 × 10⁻⁸J - ( -1.9×10⁻⁸ J)
= +7.3×10⁻⁸J.
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Answer:
Total bandwidth required = 158.2 KHz
Explanation:
given data:
number of channel 30
bandwidth of each channel is 4.5 KHz
bandwidth of guard band 0.8 KHz
According to the given information, first guard band and the guard band after last channel should be ignored, therefore we have total number of 29 guard band.
As per data, we can calculate total bandwidth required
total bandwidth = 30*4.5 + 29*0.8
total bandwidth required = 158.2 KHz
Answer:
The length of the specimen after the load is released is 11.67 cm
Explanation:
Given;
yield stress, Y = 350 MPa
ultimate tensile stress, T = 300 MPa
Elongation factor, e = yield stress, Y / ultimate tensile stress, T
Elongation factor, e = 350 Mpa / 300 Mpa
Elongation factor, e = 1.1667
New length of the specimen = 1.1667 x 10 cm = 11.67 cm
Therefore, when the load is released from 10 cm long tensile specimen, the length of the specimen becomes 11.67 cm
Answer Balanced force // Force and Motion Equal forces acting in opposite directions are called balanced forces
Explanation:
