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ivanzaharov [21]
4 years ago
15

When a pendulum is at the position all the way to the left when it is swinging (at the top of the arc), what is true of the kine

tic and potential energy? A) The potential energy is zero, and the kinetic energy is maximized. B) The kinetic energy is zero, and the potential energy is maximized. C) Both the kinetic and potential energy are zero. D) Both the kinetic and potential energy are at their maximum. D) There is no energy at this position.
Physics
1 answer:
JulsSmile [24]4 years ago
7 0
Choice - B is the correct one.

At the top of the arc, at one end of the swing:
-- it's not going to get any higher, so the potential energy is maximum
-- it stops moving for an instant, so the kinetic energy is zero

At the bottom of the arc, in the center of the swing:
-- it's not going to get any lower, so the potential energy is minimum
-- it's not going to move any faster, so the kinetic energy is maximum
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In the two-slit experiment, monochromatic light of wavelength 600 nm passes through a 19) pair of slits separated by 2.20 x 10-5
kumpel [21]

Explanation:

It is given that,

Wavelength of monochromatic light, \lambda=600\ nm=6\times 10^{-7}\ m

Slits separation, d=2.2\times 10^{-5}\ m

(a) We need to find the angle corresponding to the first bright fringe. For bright fringe the equation is given as :

d\ sin\theta=n\lambda, n = 1

\theta=sin^{-1}(\dfrac{\lambda}{d})

\theta=sin^{-1}(\dfrac{6\times 10^{-7}}{2.2\times 10^{-5}})

\theta=1.56^{\circ}

(b) We need to find the angle corresponding to the second dark fringe, n = 1

So, d\ sin\theta=(n+\dfrac{1}{2})\lambda

sin\theta=\dfrac{3\lambda}{2d}

\theta=sin^{-1}(\dfrac{3\lambda}{2d})

\theta=sin^{-1}(\dfrac{3\times 6\times 10^{-7}}{2\times 2.2\times 10^{-5}})

\theta=2.34^{\circ}

Hence, this is the required solution.

4 0
4 years ago
List the main types of electromagnetic waves in order of increasing frequency. radio waves, microwaves, infrared, visible light,
alexandr1967 [171]

Answer:

radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, gamma rays

Explanation:

Electromagnetic waves are oscillations of electric and magnetic fields in a direction perpendicular to the direction of motion of the wave (transverse waves). They are classified into 7 different types, according to their frequencies.

From lowest to highest frequency, we have:

Radio waves

Microwaves 10^9 Hz - 4\cdot 10^{13}Hz

Infrared 4\cdot 10^{13} - 4\cdot 10^{14} Hz

Visible light 4\cdot 10^{14} - 8\cdot 10^{14}Hz

Ultraviolet 8\cdot 10^{14} - 2.4\cdot 10^{16} Hz

X-rays 2.4\cdot 10^{16} -5 \cdot 10^{19}Hz

Gamma rays >5\cdot 10^{19} Hz

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Answer: 7.89141414

Explanation:

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3 years ago
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A 20.0-µF capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280
natali 33 [55]

Answer

given,

Capacitance of capacitor = 20.0-µF

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inductance = 0.280 m H

a) the oscillation frequency of circuit

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f = \dfrac{1}{2\pi}\ sqrt{\dfrac{1}{0.280 \times 10^{-3}\times 20 \times 10^{-6}}}

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b) U = \dfrac{1}{2}CV^2

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c)Current in the inductor

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3 years ago
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