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kap26 [50]
3 years ago
10

The waves from a radio station can reach a home receiver by two paths. One is a straight-line path from transmitter to home, a d

istance of 32.0 km. The second is by reflection from the ionosphere (a layer of ionized air molecules high in the atmosphere). Assume this reflection takes place at a point midway between receiver and transmitter, the wavelength broadcast by the radio station is 344 m, and no phase change occurs on reflection. Find the minimum height of the ionospheric layer that could produce destructive interference between the direct and reflected beams.
Physics
1 answer:
slava [35]3 years ago
8 0

Answer:

Explanation:

The path of waves reaching directly and through reflection from ionosphere will form a isosceles triangle.

Let the height be h . This will be height of a triangle of equal side whose base is of length 32 km. we shall calculate the length of one side of this triangle .

this length be l

l² = 16² + h²

l =  √(256 +h² )

2l = 2√(256 +h² )

path difference = 2l - 32 km

For destructive interference

path difference = wavelength /2 for minimum height .

2l - 32  = .344/2

2√(256 +h² )- 32 = . 172

2√(256 +h² ) = 32.172

4(256 +h² ) = 1035

(256 +h² ) = 258.76

h² = 2.76

h = 1.66 km

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What makes a substance, such as<br> aluminum, a good conductor?
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A tennis ball with a velocity of +10.0 m/s to the right is thrown perpendicularly at a wall. After striking the wall, the ball r
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-1500 m/s2

Explanation:

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a = \frac{\Delta v}{\Delta t} = \frac{-8 - 10}{0.012} = \frac{-18}{0.012} = -1500 m/s^2

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A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

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