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liberstina [14]
3 years ago
5

A shield of material that reflects neutrons is placed around a radioactive sample.

Physics
2 answers:
Tanya [424]3 years ago
4 0
<h3><u>Answer;</u></h3>

A) A chain reaction will be sustained in a sub-critical mass

<h3><u>Explanation</u>;</h3>
  • Critical mass is the minimum or the smallest amount of a given fissile material necessary to achieve a self-sustaining fission chain reaction under stated conditions.
  • A subcritical mass is a mass of fissile material that does not have the ability to sustain a fission chain reaction.
nikitadnepr [17]3 years ago
3 0
The answer is A. just took the test
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Explanation is given

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How do I calculate the tension in the horizontal string?
matrenka [14]

ANSWER

T₂ = 10.19N

EXPLANATION

Given:

• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,

\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

T_1\cos 30\degree-m\cdot g=0

Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

8 0
1 year ago
What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a
Alex777 [14]

Answer: d. 5 m/s^2

Explanation:

Acceleration is the change in velocity in a given time.

a = (30-20)/2 = 5

7 0
3 years ago
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A marble column of cross-sectional area 1.6 m^2 supports a mass of 26600 kg. The elastic modulus for marble is 5.0 times 10^10 N
vodka [1.7K]

Answer:

Δ L = 2.57 x 10⁻⁵ m

Explanation:

given,

cross sectional area = 1.6 m²

Mass of column = 26600 Kg

Elastic modulus, E = 5 x 10¹⁰ N/m²

height = 7.9 m

Weight of the column = 26600 x 9.8

                                    = 260680 N

we know,

Young's modulus=\dfrac{stress}{strain}

stress = \dfrac{P}{A}

           = \dfrac{260680}{1.6}

           = 162925

strain = \dfrac{\Delta L}{L}

now,

Y = \dfrac{stress}{strain}

\Delta L = \dfrac{162925}{Y}\times L

\Delta L = \dfrac{162925}{5 \times 10^10}\times 7.9

   Δ L = 2.57 x 10⁻⁵ m

The column is shortened by Δ L = 2.57 x 10⁻⁵ m

3 0
3 years ago
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