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3 years ago

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4. A hiker leaves camp and using a compass walks 4 km East, 6 km South, 3 km East, 5 km North, 10 km West, 8 km North and 3 km S

outh. What is his total displacement?

1 answer:

PSYCHO15rus [73]3 years ago

7 0

Answer:

Because the hiker walked directly west and then directly north the two legs of the hike forms a right triangle. Therefore we can use the Pythagorean Theorem to solve this problem.

c=5

The hiker is 5km from camp and should head in a generally south-east direction

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A grindstone in the shape of a solid disk with a diameter of 1.0 m and a mass of 50.0 kg, is rotating at 900 rev/min. A tool is

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Explanation:

The solubility of glucose at 30°C is

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2 years ago

You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe

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Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

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Capacitance, $C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F$

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

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So, equation (1) becomes :

$Z=\sqrt{(100)^2+(265.65-301.59)^2}$

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or

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3 years ago

Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at

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Answer:

b) twice the energy of each photon of the red light.

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Energy of a photon is given by

$E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}$

Let $\lambda_1$ = 700 nm

$\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}$

For red light

$E_1=\dfrac{hc}{\lambda_1}$

For UV light

$E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}$

Dividing the equations

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1$

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3 years ago

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Answer:

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2 years ago

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Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s

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Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density = 0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

= -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level ) = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = $\sqrt{CRT}$ = $\sqrt{1.4*287*200.5 }$ = 283.8 m/s

hence it is a subsonic aircraft

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2 years ago