Answer:
The frequency that the sampling system will generate in its output is 70 Hz
Explanation:
Given;
F = 190 Hz
Fs = 120 Hz
Output Frequency = F - nFs
When n = 1
Output Frequency = 190 - 120 = 70 Hz
Therefore, if a system samples a sinusoid of frequency 190 Hz at a rate of 120 Hz and writes the sampled signal to its output without further modification, the frequency that the sampling system will generate in its output is 70 Hz
It is study of the relationships between heat, temprature, work and energy
Answer:
<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:

<u>Where</u>:
A: is the surface area = 160
t: is the thickness = 0.002
<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>
Now, using the density we can find the mass:

Finally, with the Avogadros number (
) and with the atomic mass (A) we can find the number of atoms (N):
Hence, the number of atoms of lead required is 1.73x10²³.
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Answer:
c. and d
Explanation:
As a whistle-blower, one of your aim is to guide against unethical dealings of other people , hence you are creating an environment that uphold ethical conduct,
In addition, whistle-blowing will disclose all imminent dangers to the software community thereby preventing security breaches.
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂