Answer:
The maximum data rate supported by this line is 39900 bps
Explanation:
The maximum data rate supported by this line can be obtained using the formula below
c = W*log2(S/N+1)
where;
c is the maximum data rate supported by the line
W is the bandwidth = 4kHz
S/N+1 is the signal to noise ratio = 1001
c = 4*log2(1001)
c = 39868.9 ≅ 39900 bps
Therefore, the maximum data rate supported by this line is 39900 bps
If each mL has 30 grams of the substance in it, then 60 mL have 1800 grams of mass in them. the weight of 1,800 grams of mass on Earth is (1.8 kg) x (9.8 m/s^2) = 17.6 newtons.
We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

From the question we are told that
A solid metal ball of radius 1.5 cm
bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing
uniformly distributed charge of -7 nC
The distance between the centers of the balls is 9 cm
Generally the equation for the electric field is mathematically given as


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Answer:
145 m
Explanation:
Given:
Wavelength (λ) = 2.9 m
we know,
c = f × λ
where,
c = speed of light ; 3.0 x 10⁸ m/s
f = frequency
thus,

substituting the values in the equation we get,

f = 1.03 x 10⁸Hz
Now,
The time period (T) = 
or
T =
= 9.6 x 10⁻⁹ seconds
thus,
the time interval of one pulse = 100T = 9.6 x 10⁻⁷ s
Time between pulses = (100T×10) = 9.6 x 10⁻⁶ s
Now,
For radar to detect the object the pulse must hit the object and come back to the detector.
Hence, the shortest distance will be half the distance travelled by the pulse back and forth.
Distance = speed × time = 3 x 10^8 m/s × 9.6 x 10⁻⁷ s) = 290 m {Back and forth}
Thus, the minimum distance to target =
= 145 m
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