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3 years ago

Which nuclear reaction is used to generate electric energy in power plants?

Physics

2 answers:

mariarad [96]3 years ago

7 0

Answer: C. Nuclear fission

Explanation:

Nuclear fission is the reaction in which the nucleus of a heavy atom, when bombarded with neutrons (or when capturing an incident neutron), becomes unstable and divides into two or more nuclei of lighter atoms, giving rise to a situation of greater stability. In this process that takes place in the atomic nucleus, neutrons, gamma rays and large amounts of energy are emitted.

Then, with this division a great release of energy occurs and the emission of two or three neutrons, other particles and gamma rays.

This means fission is a process in which energy is released by the separation of the components of the nucleous of the atom.

In other words:

Matter is converted to energy, and in the case of powew plants, in electric energy.

Amiraneli [1.4K]3 years ago

4 0

Answer:

The answer is nuclear fission, like the other person said

Explanation:

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A transformer has an output coil with 6 times as many turns as the input coil. If the input frequency is f then the output frequ

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2 years ago

A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit

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A) $+1.67 rad/s^2$

The angular acceleration of the wheel is given by

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

where

$\omega_i = -6.00 rad/s$ is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

$\omega_f = 4.00 rad/s$ is the final angular velocity (anticlockwise, so with a positive sign)

$\Delta t= 6.00 s - 0=6.00 s$ is the time interval

Substituting into the equation, we find the angular acceleration:

$\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2$

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant $t_1$ where the angular velocity becomes positive (so, $\omega_i=0$ ) and the time corresponding to the final instant $t_2 = 6.0 s$ , where $\omega_f = +6.00 rad/s$ . We can find this time interval by using

$\alpha = \frac{\omega_f - \omega_i}{\Delta t}$

And solving for $\Delta t$ we find

$\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s$

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant $t_1=0$ when $\omega_i=-4.00 rad/s$ ) and the time corresponding to the instant in which the velovity becomes positive $t_2$ , when $\omega_f = 0 rad/s$ . We can find this time interval by using