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ASHA 777 [7]
2 years ago
13

Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent

ial speeds? Juan and Kuri move the same distance in one revolution, but Juan completes the revolution in less time and has a higher speed. Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed. Juan moves a shorter distance in less time than Kuri, but their speeds are the same. Juan and Kuri complete one revolution in the same time, so they have the same speed.
Physics
2 answers:
Licemer1 [7]2 years ago
6 0

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

d = 2\pi r

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

v = \frac{2\pi r}{T}

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

LuckyWell [14K]2 years ago
4 0

Answer:

B on edge

Explanation:

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Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i
Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

V = 4\frac{kQ}{r}

here

r = distance of corner of the square from it center

r = \frac{a}{\sqrt2}

r = \frac{10nm}{\sqrt2} = 7.07 nm

Q = e = -1.6 * 10^{-19} C

now the net potential is given as

V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}

V = 0.815 V

now potential energy of alpha particle at this position

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Now at the mid point of one of the side

Electrostatic potential is given as

V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}

here we know that

r_1 = \frac{a}{2} = 5 nm

r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}

r_2 = 11.2 nm

now potential is given as

V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}

V = -0.576 - 0.257 = -0.833 V

now final potential energy is given as

U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J

Now work done in this process is given as

W = U_f - U_i

W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}

W = -7 * 10^{-22} J

8 0
2 years ago
To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?
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E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.
6 0
2 years ago
If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the fr
Dmitriy789 [7]

Answer:

Frequency will be equal to 5.20 kHz

So option (c) will be correct answer

Explanation:

We have given value of capacitance C=8.5nF=8.5\times 10^{-9}f

Potential difference across capacitor V = 12 volt

Current through capacitor i=3.33mA=3.33\times 10^{-3}A

Capacitive reactance will be equal to X_c=\frac{V}{i}=\frac{12}{3.33\times 10^{-3}A}=3603.60ohm

Capacitive reactance is equal to X_c=\frac{1}{\omega C}

3603.60=\frac{1}{\omega\times  8.5\times 10^{-9}}

\omega =32647.091rad/sec

2\pi f=32647.091

f=5198.98Hz

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So frequency will be equal to 5.20 kHz

So option (c) will be correct answer

3 0
2 years ago
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tankabanditka [31]

Answer:

h= 161.06 m

Explanation:

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we know that

h=\dfrac{1}{2}gt_1^2

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Now by solving the above equation we get

h= 161.06 m

Therefore the depth of the well will be 161.06 m.

6 0
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