Answer:
the length of the pipe is 0.85 m or 85 cm
Explanation:
Given the data in the question;
The successive harmonics are; 700 Hz , 900 Hz , and 1100 H
Now, for a closed pipe,
length of pipe (L) = λ/4
Harmonics; 1x, 3x, 5x, 7x, 9x, 11x
1100Hz - 900Hz = 200Hz
⇒ 2x = 200Hz
x = 100Hz ( fundamental frequency )
λ = V/f = 340 /100 = 3.4 m
Now
Length L = λ / 4
L = 3.4 / 4
L = 0.85 m or 85 cm
Therefore, the length of the pipe is 0.85 m or 85 cm
Answer:
Explanation:
mass of displaced oil = 11 x .9
= 9.9 gm
9.9 x 10⁻³ kg
weight of displaced oil = 9.9 x 9.81 x 10⁻³ N
= .097 N .
buoyant force by oil = .097 N
weight of unknown metal = .1 x 9.8
= .98 N .
weight of metal in oil = .98 - .097
= .883 N .
=
Answer:
There would be complete destructive interference.
Explanation:
This is because since the waves are completely out of phase, the phase difference is half wavelength, that is the phase angle is 180°. The vibrating sources are 180° out of phase with each other.
Since this is the case, the crest of the one source meets the trough of the other, this causes the resultant vibrational wave to cancel out, thus producing a destructive interference pattern.
Since the vibrating sources are completely out of phase, every point they meet is completely out of phase, so the resultant interference pattern would produce a complete destructive interference pattern of no wave.
B.Electromagnetic
explanation:
Explanation:
Let us assume that the separation of plate be equal to d and the area of plates is
. As the capacitance of capacitor is given as follows.
C = 
It is known that the dielectric strength of air is as follows.
E = 
Expression for maximum potential difference is that the capacitor can with stand is as follows.
dV = E × d
And, maximum charge that can be placed on the capacitor is as follows.
Q = CV
= 
= 
= 
= 
or, = 10.62 nC
Thus, we can conclude that charge on capacitor is 10.62 nC.