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2 years ago

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A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the magnitude of the displacement (on a

straight line) of the hiker from the camp?

1 answer:

sattari [20]2 years ago

5 0

Answer:

16 km

Explanation:

Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.

The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.

11^2 + 11^2 = c^2

121 + 121 = c^2

242 = c^2

c = 15.56

Rounding the answer makes it 16 km for the hiker's magnitude of displacement.

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Read 2 more answers

The frequency of oscillation of a certain LC circuit is 299 kHz. At time t = 0, plate A of the capacitor has maximum positive ch

AnnZ [28]

Answer:

Explanation:

Given an LC circuit

Frequency of oscillation

f = 299 kHz = 299,000 Hz

AT t = 0 , the plate A has maximum positive charge

A. At t > 0, the plate again positive charge, the required time is

t =

t = 1 / f

t = 1 / 299,000

t = 0.00000334448 seconds

t = 3.34 × 10^-6 seconds

t = 3.34 μs

it will be maximum after integral cycle t' = 3.34•n μs

Where n = 1,2,3,4....

B. After every odd multiples of n, other plate will be maximum positive charge, at time equals

t" = ½(2n—1)•t

t'' = ½(2n—1) 3.34 μs

t" = (2n —1) 1.67 μs

where n = 1,2,3...

C. After every half of t,inductor have maximum magnetic field at time

t'' = ½ × t'

t''' = ½(2n—1) 1.67μs

t"' = (2n —1) 0.836 μs

where n = 1,2,3...

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2 years ago

During a test, a NATO surveillance radar system, operating at 37 GHz at 182 kW of power, attempts to detect an incoming stealth

Marta_Voda [28]

(a) $2.68\cdot 10^{-6} W/m^2$

The intensity of an electromagnetic wave is given by

$I=\frac{P}{A}$

where

P is the power

A is the area of the surface considered

For the waves in the problem,

$P=182 kW = 1.82\cdot 10^5 W$ is the power

The area is a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

so

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

So, the intensity is

$I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2$

(b) $5.9\cdot 10^{-7} W$

In this case, the area of the reflection is

$A=0.22 m^2$

So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:

$P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W$

(c) $8.7\cdot 10^{-18} W/m^2$

We said that the power of the waves reflected by the aircraft is

$P=5.9\cdot 10^{-7} W$

If we assume that the reflected waves also propagate over a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

which has an area of

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

Then the intensity of the reflected waves at the radar site will be

$I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2$

(d) $8.1\cdot 10^{-8} V/m$

The intensity of a wave is related to the maximum value of the electric field by

$I=\frac{1}{2}c\epsilon_0 E_0^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

$E_0$ is the maximum value of the electric field vector

Solving the equation for $E_0$ ,

$E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m$

(e) $1.9\cdot 10^{-16} T$

The maximum value of the magnetic field vector is given by

$B_0 = \frac{E_0}{c}$

Substituting the values,

$B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T$

And the rms value of the magnetic field is given by

$B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T$

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2 years ago