Answer:
The bulbs should be connected in parallel.
Explanation:
We want to find out a way to hook up 2 light bulbs and a battery so that when one bulb burns out or is disconnected the other bulbs stays lit.
We must connect the two bulbs in parallel so that even when one bulb is burns out, it will have no effect on the other bulb and the 2nd bulb will keep on working. The current flowing in each bulb will depend upon the resistance of each bulb and the voltage will be same across each bulb.
On the other hand, if we use a series circuit then if one bulb burns out then the there is no flow of current in the circuit and therefore, the second bulb will not be operational.
The current flowing through each bulb is given by
I = V/R
The voltage across each bulb is given by
V = IReq
Where I is the current and Req is the equivalent resistance of the two bulbs connected in parallel and is given by
Req = (R₁*R₂)/(R₁+R₂)
The connection diagram is attached where two bulbs are connected in parallel and are power with a battery.
Answer:
Sound waves travel at 343 m/s through the air and faster through liquids and solids. The waves transfer energy from the source of the sound, e.g. a drum, to its surroundings. Your ear detects sound waves when vibrating air particles cause your ear drum to vibrate. The bigger the vibrations the louder the sound.
Explanation:
It increases. Mercury takes 88 days to orbit the sun once. The Earth takes a year. Pluto takes 248 years.
Answer:
The minimum frequency is 702.22 Hz
Explanation:
The two speakers are adjusted as attached in the figure. From the given data we know that
=3m
=4m
By Pythagoras theorem
Now
The intensity at O when both speakers are on is given by
Here
- I is the intensity at O when both speakers are on which is given as 6
- I1 is the intensity of one speaker on which is 6
- δ is the Path difference which is given as
- λ is wavelength which is given as
Here
v is the speed of sound which is 320 m/s.
f is the frequency of the sound which is to be calculated.
where k=0,1,2
for minimum frequency 
, k=1
So the minimum frequency is 702.22 Hz
