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2 years ago

Same diagram... At which location would it be possible for a LUNAR ECLIPSE to happen? *

Physics

2 answers:

STALIN [3.7K]2 years ago

5 0

The answer would be a

Because a solar eclipse is where the moon covers the sun from the earth so it hat to be in front of the sunlight

Radda [10]2 years ago

5 0

A but for Lunar eclipse the shadows have to line up

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OCD would probably be the answer.

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3 years ago

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A circuit contains a single 220 pF capacitor hooked across a battery. It is desired to store three times as much energy in a com

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The capacitor should be connected in parallel as parallel connection gives the arithmetic sum of capacitance which will give a corresponding sum of energy while capacitors in series gives the sum of the reciprocal if the individual capacitance

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3 years ago

2. Light waves of the wavelength of 650 nm and 500 nm produce interference fringes on a screen at a distance of 1m from a double

satela [25.4K]

Answer:

least distance= 13mm

ratio of the lattice = 1 : 0.71 : 0.58

Explanation:

given λ₁ = 650nm = 650×10⁻⁹m, λ₂ = 500nm = 500×10⁻⁹m

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3 years ago

For a long ideal solenoid having a circular cross-section, the magnetic field strength within the solenoid is given by the equat

andrezito [222]

Answer:

Radius of the solenoid is 0.93 meters.

Explanation:

It is given that,

The magnetic field strength within the solenoid is given by the equation,

$B(t)=5t\ T$ , t is time in seconds

$\dfrac{dB}{dt}=5\ T$

The induced electric field outside the solenoid is 1.1 V/m at a distance of 2.0 m from the axis of the solenoid, x = 2 m

The electric field due to changing magnetic field is given by :

$E(2\pi x)=\dfrac{d\phi}{dt}$

x is the distance from the axis of the solenoid

$E(2\pi x)=\pi r^2\dfrac{dB}{dt}$ , r is the radius of the solenoid

$r^2=\dfrac{2xE}{(dE/dt)}$

$r^2=\dfrac{2\times 2\times 1.1}{(5)}$

r = 0.93 meters

So, the radius of the solenoid is 0.93 meters. Hence, this is the required solution.

4 0

3 years ago

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A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum s

DanielleElmas [232]

a) 1.48 m/s

The tuning fork is moving by simple harmonic motion: so, the maximum speed of the tip of the prong is related to the frequency and the amplitude by

$v_{max}=\omega A$

where

$v_{max}$ is the maximum speed

$\omega$ is the angular frequency

A is the amplitude

For the tuning fork in the problem, we have

$\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s$ , where f is the frequency

$A=0.600 mm=6\cdot 10^{-4} m$ is the amplitude

Therefore, the maximum speed is

$v_{max}=(2462 rad/s)(6\cdot 10^{-4}m)=1.48 m/s$

b) $3.0\cdot 10^{-5} J$

The fly's maximum kinetic energy is given by

$K=\frac{1}{2}mv_{max}^2$

where

$m=0.0270 g=2.7\cdot 10^{-5} kg$ is the mass of the fly

$v_{max}=1.48 m/s$ is the maximum speed

Substituting into the equation, we find

$K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J$

7 0

3 years ago