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coldgirl [10]
3 years ago
8

Who's really good at physical science?

Physics
1 answer:
34kurt3 years ago
3 0
Definitely not me!!!! (unless it's magnetism or circuits)
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Energy is released during the fission of pu-239 atoms as a result of the
pav-90 [236]
<h2>Answer: process of converting matter into energy</h2><h2></h2>

Nuclear fission consists of dividing a heavy nucleus into two or more lighter or smaller nuclei, by means of the bombardment with neutrons to make it unstable. In this process that takes place in the atomic nucleus, neutrons, gamma rays and <u>large amounts of energy are emitted.  </u>

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:

<h2>Matter is converted to energy .</h2>

6 0
3 years ago
What would a series circuit be used for?
igor_vitrenko [27]

Answer:

C

Explanation:

a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer

6 0
3 years ago
A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn
jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = \sqrt{Cx^{2}+Cy^{2}  }

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

brainly.com/question/3184914

#SPJ1

6 0
2 years ago
How many resistors are found in this circuit? <br> A) 0<br> B) 1<br> C) 2<br> D) 3
Dmitriy789 [7]
Show us the pictures I don't see it
6 0
3 years ago
A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.
Xelga [282]

Answer:

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

Explanation:

As we know that magnetic field due to torroid is given as

B = \frac{\mu_0 N i}{2\pi b}

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

\phi = B.A

\phi = \frac{\mu_0 N i}{2\pi b}(a^2)

now total flux of N number of coils is given as

\phi_{total} = \frac{\mu_0 N^2 i(a^2)}{2\pi b}

now we know that self inductance is the property of coil in which flux of the coil will link with the current in the coil

So we know that

L = \frac{\phi}{i}

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

3 0
3 years ago
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