Answer:
Hi myself Shrushtee.
Explanation:
The fuse is connected to the live wire so that the appliance will not become charged (have a potential difference of 230 V) after the fuse has melted due to excessive current. Fuses must be fitted onto the live wire so that when it blows, it will disconnect (isolate) the appliance from the high voltage live wire.
Complete Question
A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.
(a)
What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding
(b) What wavelength is detected by a person on the platform as the train approaches?
Answer:
a

b

Explanation:
From the question we are told that
The speed of the train is 
The frequency of the train horn is 
Generally the speed of sound has a constant values of 
Now according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

substituting values


Now according to dopplers equation when the train(source) moves away from the person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as

substituting values


The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as



Generally the wavelength detected by the person as the train approaches is mathematically represented as



Answer:
a

b

Explanation:
From the question we are told that
The maximum angular speed is 
The time taken is 
The minimum angular speed is
this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that

=> 
substituting values


converting to 
We have


According to the first equation of motion the angular displacement is mathematically represented as

substituting values


converting to revolutions


The source on the nagnetic field on the ssun is the moons pull
Answer:
Part a)

so here the angle made by the string is independent of the mass
Part b)

Explanation:
Part a)
Let the string makes some angle with the vertical so we have force equation given as


so we will have


so here the angle made by the string is independent of the mass
Part b)
Now from above equation if we know that angle made by the string is

so we will have



