for this we apply, Heisenberg's uncertainty principle.
it states that physical variables like position and momentum, can never simultaneously know both variables at the same moment.
the formula is,
Δp * Δx = h/4π
m(e).Δv * Δx = h/4π
by rearranging,
Δx = h / 4π * m(e).Δv
Δx = (6.63*10^-34) / 4 * 3.142 * 9.11*10^-31 * 5.10*10^-2
Δx = 6.63*10^-34 / 583.9 X 10 ⁻³¹
Δx = 0.011 X 10⁻³
for the bullet
Δx = (6.63*10^-34) / 4 * 3.142 * 0.032*10^-31 * 5.10*10^-2
Δx = 6.63*10^-34 /2.05
Δx =3.23 X 10⁻³² m
therefore, we can say that the lower limits are 0.011 X 10⁻³ m for the electron and 3.23 X 10⁻³² m for the bullet
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Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 ×
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 ×
)²×ω²× 34.64 ) / 2
0.050 ×(6.43 ×
)²×ω²× 34.64 = 108
ω² = 108 / 7.160 ×
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
Answer:
the extension recorded by the student would be smaller than the actual extension of the spring
Answer:
I believe the answer to be B
Explanation:
If you were playing on grass, the ball would be able to roll around much easier rather than it to be on sand. If it's wrong I am so sorry
D = 1/2 g t^2. It works out to 44.1 meters.