Uranium-235 Fission

1 answer:

4 0

In the balanced nuclear reaction of uranium 235, we have thorium 231, helium atom and energy released at the right of the equation

<h3>Balanced nuclear reaction of Uranium</h3>

The balanced nuclear reaction of uranium is determined as follows;

$^{235}_{92}U\ --- > \ ^4_2He\ + \ ^{231}_{90}Th\ + \ energy$

Thus, in the balanced nuclear reaction of uranium 235, we have thorium 231, helium atom and energy released at the right of the equation while uranium 235 is on the left hand side of the nuclear reaction equation.

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A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work

kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

3 0

3 years ago

A bar of silicon is 4 cm long with a circular cross section. If the resistance of the bar is 270 ω at room temperature, what is

vova2212 [387]

solution:

$consider the following data\\
length of slicon bar with circular cross section is 4cm or 0.04m\\
at room temperature resistance of the slicon bar is 270\Omega \\
represent the resistance in mathematical from\\
r=p\frac{1}{A}---1\\
where r is resistance and l is the length \\
A is cross sectional area\\
it is clear that resistivity of the silicon meterial is 6.4\times^2 \Omega.m\\
substitute 6.4\times10^2 for p,270\Omega for R and 0.04m for l i equation (1).\\$ $270=(604\times10^2)\frac{0.04}{A}\\
rewrite the equation\\
a=(6.4\times10^2)\frac{(0.04)}{270}\\
=0.9481m^2\\
write the formula for the circular cross sectional area of silicon bar.\\
A=\pi r^2\\
substitute 0.9481 for A in the above equation\\
\pi r^2=0.9481
r^2=\frac{0.9481}{3.14},since \pi =3.14\\
0.30194\\
further simplified\\
r^2=0.30194\\
\sqrt{0.30194}\\
\cong 0.1509m\\
\cong 150.1mm$