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Molodets [167]
3 years ago
12

If the maximum time between the emitted and received pulse is 1/ 10 second, what is the farthest distance you could measure with

the sensor? (This question requires units.)
Physics
1 answer:
Allisa [31]3 years ago
7 0

Answer:

d = 2,998 10⁷ m

Explanation:

The speed of an electromagnetic wave is the speed of light

      c = 2,998 10⁸ m / s

This speed is in a vacuum, when the light is in a material medium, its speed is lower than this, even when the order of magnitude is maintained.

Since the speed of a wave is constant, we can use the kinematic relationship

       c = d / t

       d = c t

Let's calculate

      d = 2,998 108 0.1

      d = 2,998 10⁷ m

in the vacuum or air

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99% of all the matter that can be observed in the universe exists as
Temka [501]
Has mass and occupied space
5 0
4 years ago
Two long, parallel transmission lines, 40.0cm apart, carry 25.0-A and 73.0-A currents.A). Find all locations where the net magne
In-s [12.5K]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

5 0
4 years ago
An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and
Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is F_{ma}=BqV---(i)

The given formula to find the electric force is F_{el}=qE---(ii)

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

Bqv=qE\\\\v=\frac{E}{B}

v=\frac{187500}{0.125} \\\\v=15\times10^5m/s

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

F_{ce}=\frac{mv^2}{r}---(iii)

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br}  \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg

Therefore,  the particle's charge-to-mass ratio is 958.1\times10^5C/kg

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0
4 years ago
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r
pogonyaev

Given:

L = 1 mH = 1\times 10^{-3} H

total Resistance, R = 11 \Omega

current at t = 0 s,

I_{o} = 2.8 A

Formula used:

I = I_{o}\times e^-{\frac{R}{L}t}

Solution:

Using the given formula:

current after t = 0.5 ms = 0.5\times 10^{-3} s

for the inductive circuit:

I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}

I =   2.8\times e^-5.5

I =0.011 A

5 0
3 years ago
What is one suggestion to help ensure a child gets adequate sleep?
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Start a bedtime routine.
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6 0
3 years ago
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