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Anarel [89]
3 years ago
13

What is the volume of a cone with a height of 27cm and a radius of 13cm? And your answer to the nearest tenth

Physics
1 answer:
SashulF [63]3 years ago
5 0

Answer:

the volume of the cone is 4778.36 cm^3

Explanation:

The computation of the volume of the cone is shown below:

As we know that

The Volume of the cone is

r denotes the radius

And, h denotes the height

So,

= \frac{1}{3} \pi r^2 h \\\\=  \frac{1}{3} \times 3.14 \times 13^2 \times 27\\\\= 4778.36 cm^3

Hence, the volume of the cone is 4778.36 cm^3

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A baseball player hits a ball with a bat. At one moment the force exerted by the bat on the ball is 18,000 N. At that same momen
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Answer:

Less than 18000N

Explanation:

Given

Force\ Exerted = 18000N

This question will be answered using Newton's third law.

Understanding this law, it implies that reaction force is equal and opposite to the force exerted.

This implies that;

If the force exerted on the ball is 18000N

the force exerted is -18000N

So, the option that answers the question is less than 18000N because -18000N < 18000N

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rocks and other materials inside the Earth. 5.

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A large, massive object collides with a stationary, smaller object on an ice rink. If the large object transfers all of its mome
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An early model of the atom, proposed by Rutherford after his discovery of the atomicnucleus, had a positive point charge +Ze(the
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Answer:

a)  E = k Ze (1- r³ / R³)  1/r², b) E=0, c)   E = -6.62 10¹⁰  N / C

Explanation:

a) For this we can use the law of Gauus

         Ф = E- dA = q_{int} / ε₀

where we take a sphere as a Gaussian surface, so that the electric field lines and the radii of the sphere are parallel, consequently the dot product is reduced to the algebraic product

       E A =q_{int} / ε₀

the area of ​​a sphere  

      A = 4π r²

      E 4π r² = q_{int} / ε₀

      E = 1 / 4πε₀   q_{int} / r²

       k = 1 /4π ε₀

       E = k q_{int} / r²       (1)

       

let's analyze the charge inside the gaussian sphere,

let's use the concept of density for electrons, since they indicate that the charge is evenly distributed

     ρ = Q / V

where the volume of the sphere is

    V = 4/3 πr³

     Qe = ρ V

     Qe = ρ 4 / 3π r³

the density of the electrons is

     ρ = Ze 3 / (4π R³)

where R is the atomic radius

we substitute

       Qe = Ze r³/ R³

for protons they are in a very small space, the atomic nucleus, so we can superno that they are a point charge.

The net charge inside our Gaussian surface, the charge of the protoens plus the charge of the electroens (Qe)

     q_{int} = q_proton + Q_electron

     q_{int} = + Ze - Qe

     q_{int} = + Ze - Ze r³ / R³

     q_{int} = Ze (1- r³ / R³)

   

  we substitute in equation 1

     E = k Ze (1- r³ / R³)  1/r²

b) on the surface of the atom r = R

therefore the electric field is zero

      E = 0

c) Calculate the electric field for the Uranium for

       r = R / 2 = 0.10 10⁻⁹ / 2 = 0.05 10⁻⁹ = 5 10⁻¹¹ m

     

       E = 8.99 10⁹ 92 1.6 10⁻¹⁹ (1-1/2)³   1/ (5 10⁻¹¹)²

       E = -6.62 10¹⁰  N / C

6 0
4 years ago
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