Answer:
The resonant frequency of this circuit is 14.5 kHz.
Explanation:
Given that,
Inductance of a parallel LCR circuit, 
Capacitance of parallel LCR circuit, 
At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :
or
f = 14.5 kHz
So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.
Answer:
Explanation:
As spring season is a yearly phenomenon so, Rita should organize her data on yearly basis. Firstly, she should plan the procedure of her experiment and collect the data according to it. Secondly, identify the attribute of each object of her experiment. Thirdly, she can organize and segregate her data in tabular form, graphical form or diagrammatically.
Winds are named based on which compass direction the wind is blowing. For example some common ones are NE or N or SE or SW. NE stands for Northeast, N for North, SE for South East and SW for Southwest.
Answer:
Explanation:
Given
Lowest four resonance frequencies are given with magnitude
50,100,150 and 200 Hz
The frequency of vibrating string is given by
where n=1,2,3 or ...n
L=Length of string
T=Tension
Mass per unit length
When string is clamped at mid-point
Effecting length becomes 
Thus new Frequency becomes
i.e. New frequency is double of old
so new lowest four resonant frequencies are 100,200,300 and 400 Hz
Answer:
a) 4.31 m/s²
b) 215.5 m
Explanation:
a) According to Newton's first law of motion
The net force applied to particular mass produced acceleration, a, according to
F = ma
F = 140 N
m = 32.5 kg
a = ?
140 = 32.5 × a
a = 140/32.5 = 4.31 m/s²
b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s
u = initial velocity of the probe = 0 m/s (since it was initially at rest)
a = 4.31 m/s²
t = 10 s
s = distance travelled = ?
s = ut + at²/2
s = 0 + (4.31×10²)/2 = 215.5 m
