Answer:
0.0389 cm
Explanation:
The current density in a conductive wire is given by
where
I is the current
A is the cross-sectional area of the wire
In this problem, we know that:
- The fuse melts when the current density reaches a value of
- The maximum limit of the current in the wire must be
I = 0.62 A
Therefore, we can find the cross-sectional area that the wire should have:
We know that the cross-sectional area can be written as
where d is the diameter of the wire.
Re-arranging the equation, we find the diameter of the wire:
A) the pump removes air from the enclosure. foam absorbs vibrations from the clock
b) result is that as the air is taken out by pump, the sound of alarm reduces and eventually becomes zero. so, you cant hear the alarm ringing.
c) conclusion is that sound cant travel in vaccum and it requires a medium to travel.
The second ball should strike at double the original t value
Answer:
a. Wavelength = λ = 20 cm
b. Next distance of maximum intensity will be 40 cm
Explanation:
a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.
Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.
This we get the equation:
(nλ + λ/2) - nλ = 30-20
λ/2 = 10
λ = 20 cm
b. at what distance, sound intensity will be maximum again.
For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.
= 30 + λ/2
= 30 + 20/2
=30+10
=40 cm
