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uysha [10]
3 years ago
5

Which of the following is a typical product of nuclear fission?

Physics
2 answers:
xxMikexx [17]3 years ago
6 0
The correct answer for the question that is being presented above is this one: "c. radiation." The typical product of nuclear fission is radiation. Nuclear fission's another product is that<span> two different atoms that are both less massive than the original atom </span>
krok68 [10]3 years ago
6 0

On edge.nuity the answer is C- radiation. Hope this helps you out :)

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A piece of warm concrete is placed in a cold-water tank, and energy flows between the concrete and the water. Which way does the
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Answer:

Heat flows from hot to cold objects. When a hot and a cold body are in thermal contact, they exchange heat energy until they reach thermal equilibrium, with the hot body cooling down and the cold body warming up. This is a natural phenomenon we experience all the time.

Explanation:

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Calculate the kinetic energy of a 4.00kg object traveling at 16.0 m/s
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KE = ½mv² = ½(4.00 kg)(16.0 m/s)² = 512 J
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Young's Modulus refers to changes in the a Volume b- Length c- Body layers
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Explanation:

Young' modulus is the ratio of normal stress to the longitudinal strain. Mathematically, it is given by :

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Y=\dfrac{F/A}{\Delta L/L}..........(1)

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3 years ago
Which type of mixture could this illustration represent?
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The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
Vedmedyk [2.9K]

Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

Coefficient of static friction is, \mu = 0.214

Angle of inclination is, \theta = 31.5°

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

Therefore, the  tension in the rope is 229.37 N.

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