Which sentence best describes the axis of rotation of Earth?

Two antennas located at points A and B are broadcasting radio waves of frequency 96.0 MHz, perfectly in phase with each other. T

earnstyle [38]

Answer:

a) Δφ = 1.51 rad , b) x = 21.17 m

Explanation:

This is an interference problem, as they indicate that the distance AP is on the x-axis the antennas must be on the y-axis, the phase difference is

Δr /λ = Δfi / 2π

Δfi = Δr /λ 2π

Δr = r₂-r₁

let's look the distances

r₁ = 57.0 m

We use Pythagoras' theorem for the other distance

r₂ = √ (x² + y²)

r₂ = √(57² + 9.3²)

r₂ = 57.75 m

The difference is

Δr = 57.75 - 57.0

Δr = 0.75 m

Let's look for the wavelength

c = λ f

λ = c / f

λ = 3 10⁸ / 96.0 10⁶

λ = 3.12 m

Let's calculate

Δφ = 0.75 / 3.12 2π

Δφ = 1.51 rad

b) for destructive interference the path difference must be λ/2, the equation for destructive interference with φ = π remains

Δr = (2n + 1) λ / 2

For the first interference n = 0

Δr = λ / 2

Δr = r₂ - r₁

We substitute the values

√ (x² + y²) - x = 3.12 / 2

Let's solve for distance x

√ (x² + y²) = 1.56 + x

x² + y² = (1.56 + x)²

x² + y² = 1.56² + 2 1.56 x + x²

y2 = 20.4336 +3.12 x

x = (y² -20.4336) /3.12

x = (9.3² -20.4336) /3.12

x = 21.17 m

This is the distance for the first minimum

6 0

3 years ago

Read 2 more answers

What effect does the Boat Velocity have on the waves seen by the observer?

photoshop1234 [79]

sbjananaan swhhwvwwhehs

5 0

3 years ago

A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle

denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/ $s^{2}$ = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = $\frac{8.04N}{70.2Kg} = 0.1130801680m/s^2$ .

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0

3 years ago

Which two statements are true about a system?

shusha [124]

A and c

Because it defines the boundaries of the system

4 0

3 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago