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3 years ago

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A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.1 m/s. The car is a distanc

e d away. The bear is 29 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?

2 answers:

Igoryamba3 years ago

6 0

It's 61.5m

by solving simultaneously equations
29+d = 6t (bear)
d=4.1t (tourist)

vodomira [7]3 years ago

5 0

<span>A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.1 m/s. The car is a distance d away. The bear is 29 m behind the tourist and running at 6.0 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
</span>
61.5m

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Answer:

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Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

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3 years ago

A compound microscope has an objective lens of focal length 1.40 cm and an eyepiece with a focal length of 2.20 cm. The objectiv

olchik [2.2K]

Answer:

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3 years ago

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

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(a) $7.32\cdot 10^7 Hz$

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E = 310 V/m

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$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

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- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

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$I=\frac{E^2}{2\mu_0 c}$

where we have

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Substituting into the formula,

$I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2$

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$\frac{dp}{dt}=\frac{A}{c}$

where the <S> is the magnitude of the Poynting vector, given by

$=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2$

and where the surface is

A = 1.8 m^2

Substituting, we find

$\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s$

(h) $8.47\cdot 10^{-7} N/m^2$

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$p=\frac{}{c}$

where we have

$=254.2 W/m^2$

$c=3\cdot 10^8 m/s$

Substituting, we find

$p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2$

8 0

3 years ago