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3 years ago

What is the minimum total energy released when

Physics

2 answers:

Karo-lina-s [1.5K]3 years ago

7 0

Answer:

(1) $1.64 \times 10^{-13} J$

Explanation:

Energy released in this process is given by

$Q = \Delta m c^2$

here we have

$\Delta m = 9.1 \times 10^{-31} kg$

since here mass of two electrons converted into energy so we have

$Q = 2(9.1 /times 10^{-31})(3\times 10^8)^2$

$Q = 1.64 \times 10^{-13} J$

so here energy released is the energy of rest mass energy due to two charges i.e. electrons and positrons

Ganezh [65]3 years ago

6 0

What is the minimum total energy released when an electron and its antiparticle (positron) <span>annihilate each other?
The </span>minimum total energy released when an electron and its antiparticle (positron) annihilate each other is <span>2.73 × 10^–22 J. The answer is number 4.</span>

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What is the focal length of a lens

goldenfox [79]

Answer:

The focal length of a lens is refers to the distance from the center of the lens to the principal foci.

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3 years ago

Discuss the force that exists between the Earth and the moon by referring to the mass of each.

Fudgin [204]

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Hope it answers your question!

Brainliest would be nice but of course you don’t gotta :)

7 0

2 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

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3 years ago

Assume that the loop is initially positioned at θ=30∘θ=30∘ and the current flowing into the loop is 0.500 AA . If the magnitude

labwork [276]

Answer: $\tau=1.03\times 10^{-4}\ N-m$

Torque,

Explanation:

Given that,

The loop is positioned at an angle of 30 degrees.

Current in the loop, I = 0.5 A

The magnitude of the magnetic field is 0.300 T, B = 0.3 T

We need to find the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field. We know that the torque is given by :

$\tau=NIAB\ \sin\theta$