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statuscvo [17]
2 years ago
15

2. Find rectangular coordinates for the point (6, 240°).

Physics
1 answer:
tatuchka [14]2 years ago
7 0

Answer:

(-3, -3√3)

Explanation:

To convert from polar coordinate to rectangular coordinate, first you have to know these two equations:

\displaystyle{x=r\cos \theta}\\\displaystyle{y=r\sin \theta}

We know that (x,y) is in rectangular coordinate form while (r,θ) is in polar coordinate form.

Therefore, substitute r = 6 and θ = 240° in both equations:

\displaystyle{x=6\cos 240^{\circ}}\\\displaystyle{y=6\sin 240^{\circ}}

After evaluating, you'll get:

\displaystyle{x=-3}\\\displaystye{y=-3\sqrt{3}}

Therefore, substitute x and y in rectangular coordinate form - hence, the answer is (-3, -3√3)

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What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?
zubka84 [21]

Answer:

vp = 0.94 m/s

Explanation

Formula

Vp = position/ time

position: Initial position - Final position

Position = 25 m - (-7 m) = 25 m + 7 m = 32 m

Then

Vp = 32 m / 34 seconds

Vp = 0.94 m/s

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Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.
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Answer:

4.29 millimeters

Explanation:

Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

v=343 m/s

The frequency of the waves emitted by this bat is:

f=80.0 kHz = 80,000 Hz

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

\lambda=\frac{v}{f}=\frac{343 m/s}{80,000 Hz}=4.29\cdot 10^{-3} m=4.29 mm

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According to the Big Bang Theory, how long ago did the universe expand explosively into existence
Dmitry_Shevchenko [17]

About 13.7 billion years ago

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The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc
Ivanshal [37]

1. Frequency: 3.23\cdot 10^{20} Hz

The energy given is the energy per mole of particles:

E=1.29\cdot 10^{11} J/mol

1 mole contains a number of Avogadro of particles, N_A, equal to

N_A=6.022\cdot 10^{23} particles

So, by setting the following proportion, we can calculate the energy of a single photon:

1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J

This is the energy of a single photon; now we can calculate its frequency by using the formula:

E_1 = hf

where

h=6.63\cdot 10^{-34} Js is the Planck's constant

f is the photon frequency

Solving for f, we find

f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz

2. Wavelength: 9.29\cdot 10^{-13} m

The wavelength of the photon is given by the equation:

\lambda=\frac{c}{f}

where

c=3\cdot 10^8 m/s

is the speed of the photon (the speed of light). Substituting,

\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m

6 0
3 years ago
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