Hiran is standing beside the road when he hears a bird flying away from hip and chirping. The bird’s chirp has a frequency of 18
1 answer:
The frequency of bird chirping hear by hiran will be 1.77 kHz.
<u>Explanation:</u>
As per Doppler effect, the observer will feel a decrease in the frequency of the receiving signal if the source is moving away from the observer. So the shifted frequency is obtained using the below equation:
Here , c is the speed of sound, Vs is the velocity of source with which it is moving away. f is the original frequency of source and f' is the frequency shift heard by the observer.
As here, f = 1800 Hz, Vs= 6 m/s and c = 343 m/s, then
So, the frequency of bird chirping hear by hiran will be 1.77 kHz.
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Answer:
2.86 m
Explanation:
Given:
M₁ = 10 kg
M₂ = 5 kg
= 0.5
height, h = 5 m
distance traveled, s = 2 m
spring constant, k = 250 N/m
now,
the initial velocity of the first block as it approaches the second block
u₁ = √(2 × g × h)
or
u₁ = √(2 × 9.8 × 5)
or
u₁ = 9.89 m/s
let the velocity of second ball be v₂
now from the conservation of momentum, we have
M₁ × u₁ = M₂ × v₂
on substituting the values, we get
10 × 9.89 = 5 × v₂
or
v₂ = 19.79 m/s
now,
let the velocity of mass 2 when it reaches the spring be v₃
from the work energy theorem, we have
Work done by the friction force = change in kinetic energy of the mass 2
or
or
v₃ = 20.27 m/s
now, let the spring is compressed by the distance 'x'
therefore, from the conservation of energy
we have
Energy of the spring = Kinetic energy of the mass 2
or
on substituting the values, we get
or
x = 2.86 m
Answer:
Decrease
Explanation:
I = Current = 3.7 A
e = Charge of electron =
n = Conduction electron density in copper =
= Drift velocity of electrons
r = Radius = 1.23 mm
Current is given by
The drift speed of the electrons is
From the equation we can see the following
So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.