Answer:
The frequency of the tuning is 1.065 kHz
Explanation:
Given that,
Length of tube = 40 cm
We need to calculate the difference between each of the lengths
Using formula for length
For an open-open tube,
We need to calculate the fundamental wavelength
Using formula of wavelength
Put the value into the formula
We need to calculate the frequency of the tuning
Using formula of frequency
Put the value into the formula
Hence, The frequency of the tuning is 1.065 kHz
Solution :
The given figure is a loop of a wire with a resistor.
When the switch S is closed for long time and is suddenly opened, the direction of the induced current can be find out by using the rule of right hand screw. According to the right hand screw rule, the direction of the magnetic field at the loop is in the direction that points outwards. The strength of the current rapidly decreases as it is switch off and the magnetic flux that is linked with the loop wire will also decrease.
According to the Lenz's law, the direction of the induced current must be such the decrease in the magnetic flux. It means the direction of the magnetic field must be outwards and also normal to the plane of the screen. The direction of the induced anti clockwise or from right to left in the resistance.
Answer:
(a) Electric field at 0.250 m is zero.
(b) Electric field at 2.90 m is zero.
(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.
(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.
Explanation:
Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.
Electric field inside and outside the metal sphere is :
E = 0 for r ≤ R ( inside )
= for r > R ( outside )
Here K is electric constant and r is the distance from the center of the metal sphere.
(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.
(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.
(c) Electric field at 3.10 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.
E = -
E = - 5.15 x 10³ V/m
(d) Electric field at 8.00 m is given by the relation as r > R :
E =
Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.
E = -
E = - 0.77 x 10³ V/m