Answer:
other part will move with momentum 0.5 kg m/s towards west
Explanation:
As we know that the during the explosion event the time interval is very small and there is no impulsive force on the firecracker during this time
so here we can say that the momentum will remains conserved during the this event
now if the initial momentum of the firecracker is zero
then the final momentum must be zero
so here we can say
towards east
now from above equation
so the other part will move with momentum 0.5 kg m/s towards west
Answer:
The velocity of the wave is 12.5 m/s
Explanation:
The given parameters are;
he frequency of the tuning fork, f = 250 Hz
The distance between successive crests of the wave formed, λ = 5 cm = 0.05 m
The velocity of a wave, v = f × λ
Where;
f = The frequency of the wave
λ = The wavelength of the wave - The distance between crests =
Substituting the known values gives;
v = 250 Hz × 0.05 m = 12.5 m/s
The velocity of the wave, v = 12.5 m/s.
Answer: T = 472.71 N
Explanation: The wire vibrates thus making sound waves in the tube.
The frequency of sound wave on the string equals frequency of sound wave in the tube.
L= Length of wire = 26cm = 0.26m
u=linear density of wire = 20g/m = 0.02kg/m
Length of open close tube = 86cm = 0.86m
Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as
L = 3λ/4
0.86 = 3λ/4
3λ = 4 * 0.86
3λ = 3.44
λ = 3.44/3 = 1.15m.
Speed of sound in the tube = 340 m/s
Hence to get frequency of sound, we use the formulae below.
v = fλ
340 = f * 1.15
f = 340/ 1.15
f = 295.65Hz.
f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.
The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as
L = λ/2
0.26 = λ/2
λ = 0.52m
The speed of sound in string is given as v = fλ
Where λ = 0.52m f = 295.65 Hz
v = 295.65 * 0.52
v = 153.738 m/s.
The velocity of sound in the string is related to tension, linear density and tension is given below as
v = √(T/u)
153.738 = √T/ 0.02
By squaring both sides
153.738² = T / 0.02
T = 153.738² * 0.02
T = 23,635.372 * 0.02
T= 472.71 N
-- 400 nm shifted to 430 nm . . . longer than it should be; "red shifted"; moving away from Earth
-- 610 nm shifted to 580 nm . . . shorter than at source; "blue shifted"; moving toward Earth
-- 512 nm shifted to 480 nm . . . shorter than at source; moving toward Earth
-- 670 nm shifted to 690 nm . . .longer than at source; moving away from Earth
Now I'd just like to ask one more itty bitty question, that you can think about while you're on this subject: Astronomers really do this. They measure how much the wavelength CHANGED, from the time it left the original source until the time they detect it. But HOW do they know what the wavelength WAS when it left the source ? ? ?
THIS is the part that blows my mind !
1) Magnetic field
2) Electric current
3) Force
