If the delay of the 10-bit adder is 1ns, the setup time, the hold time, and the propagation delay of the registers is 0.3 ns, 0.
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Missing detail in the text:
"<span>A small glass bead has been charged to + 25 nC "
Solution
The force exerted on a charge q by an electric field E is given by
</span>
<span>Considering the charge on the bead as a single point charge, the electric field generated by it is
</span>
with
, 
is the charge on the bead. We want to calculate the field at 
:
The proton has a charge of
, therefore the force exerted on it is
And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass,
, we have
The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so:
"<span>A small glass bead has been charged to + 25 nC "
Solution
The force exerted on a charge q by an electric field E is given by
</span>
<span>Considering the charge on the bead as a single point charge, the electric field generated by it is
</span>
with
The proton has a charge of
And finally, we can use Newton's second law to calculate the acceleration of the proton. Given the proton mass,
The charge on the bead is positive, and the proton charge is positive as well, therefore the proton is pushed away from the bead, so: