) The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is 0.23
1 answer:
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Answer:
The spring stretched by x = 13.7 cm
Explanation:
Given data
Mass = 3 kg
k = 120
Angle = 34°
From the free body diagram
Force acting on the box = mg sin
⇒ F = 3 × 9.81 ×
⇒ F = 16.45 N ------- (1)
Since box is attached with the spring so a spring force also acts on the box.
= k x
= 120
-------- (2)
The net force acting on the body is given by
Since acceleration of the box is zero so
Put the values from equation (1) & (2) we get
16.45 = 120
x = 0.137 m
x = 13.7 cm
Therefore the spring stretched by x = 13.7 cm
Explanation:
Given that,
Distance 1, r = 100 m
Intensity,
If distance 2, r' = 25 m
We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :
.........(1)
Let I' is the intensity at r'. So,
............(2)
From equation (1) and (2) :
Intensity level is given by :
,
dB = 32.96 dB
Hence, this is the required solution.