Answer:
Explanation:
From the question we are told that
Distance b/e antenna's 
Frequency of antenna Radiation
Distance from receiver 
Intensity of Receiver 
Distance difference of the receiver b/w antenna's 
Generally the equation for Phase difference 
is mathematically given by
<h3>
</h3>
Therefore phase difference f between the two radio waves produced by this path difference is given as
Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>. That description is a little more direct.
It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".
Answer:
Explanation:
We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .
f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .
f will be lowest when v₀ is highest .
velocity of observer is highest when he is at the equilibrium position or at middle point .
So apparent frequency is lowest when observer is at the middle point and going away from the source while swinging to and from before the source of sound .
The momentum of the
x-ray photon is p = h/lambda . Lambda is the wavelength (0.30nm=3x10^(-9)m) and
h is Planck's constant,(h=6.62607004 × 10-34<span> m2 kg / s).The
momentum is: 2.2 x 10^(-25).</span>
The momentum can be calculated
also as: p=mv, where m is the mass of the electron and v is the speed.
So v=p/m,p is known,and
also the mass of the electron (m=9.10938356 × 10-31<span> kilograms).</span>
v=2.2 x 10^(-25)/9.10938356
× 10-31<span> kilograms=0.24 x 10^6 m/s</span>
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S
= 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S > σ
