Four capacitors with capacitance 3.0 pF, 2.0 pF, 5.0 pF and X pF are connected in series to each other. If the equivalent capaci
1 answer:
Answer:
<h2>A. 6pF</h2>Explanation:
If unknown capacitance C1, C2, C3 and C4 are connected in series to one another, their equivalent capacitance of the circuit will be expressed as shown
Given the capacitance's 3.0 pF, 2.0 pF, 5.0 pF and X pF connected in series to each other. If the equivalent capacitance of the circuit is 0.83 pF, then to get X, we will apply the formula above;
C₄ ≈ 6pF
Hence the value of the X capacitor is approximately 6pF
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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
Answer:
19.5°
Explanation:
The energy of the mass must be conserved. The energy is given by:
1)
where m is the mass, v is the velocity and h is the hight of the mass.
Let the height at the lowest point of the be h=0, the energy of the mass will be:
2)
The energy when the mass comes to a stop will be:
3)
Setting equations 2 and 3 equal and solving for height h will give:
4)
The angle ∅ of the string with the vertical with the mass at the highest point will be given by:
5)
where l is the lenght of the string.
Combining equations 4 and 5 and solving for ∅:
6)