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Ne4ueva [31]
3 years ago
5

Three heavy rods are all made of the same uniform material. These rods have lengths 3 m, 4 m, and 5 m, and compose the sides of

a 3-4-5 right triangle. Find the coordinates of the center of mass of the triangle. (Assume that the right angle is at the origin, and that the two perpendicular sides are parallel to the axes, with the longer leg in the vertical direction.)
Physics
1 answer:
Sonbull [250]3 years ago
3 0

Answer:

[ 2.67 , 1 ] m

Explanation:

Given:-

- The side lengths of the rods are as follows:

                             a = 4 m , b = 4 m , c = 5 m

                             a = Base , b = Perpendicular , c = Hypotenuse

- All rods are made of same material with uniform density. With  

Find:-

Find the coordinates of the center of mass of the triangle.

Solution:-

- The center of mass of any triangle is at the intersection of its medians.

- So let’s say we have a triangle with vertices at points (0,0) , (a,0) , and (0,b).

  • Median from (0,0) to midpoint (a/2,b/2) of opposite side has equation:

                                       bx−ay=0

  • Median from (a,0) to midpoint (0,b/2) of opposite side has equation:

                                      bx+2ay=ab

  • Median from (0,b) to midpoint (a/2,0) of opposite side has equation:

                                     2bx+ay=ab

  • Solve all three equations simultaneously:

                                     bx−ay=0  , bx = ay

                                     ay + 2ay = ab , 3ay = ab , y = b/3

                                     bx = b/3

                                     x = a / 3

  • So the distance from the median to each leg of the triangle is 1/3 length of other leg.

- So the coordinates of the centroid for right angle triangle would be:

                                   [ 2a/3 , b/3 ]

                                   [ 2.67 , 1 ] m

                                                         

                                 

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Rashid [163]

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

            λ = 102.78  nm

This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

5 0
3 years ago
A hockey puck is traveling to the left with a velocity of v=10 when it is struck by the hockey stick
Lera25 [3.4K]
We have to calculate the impulse of a hockey puck.
Imp = m * ( v 1 - v 2 ) = m * Δ v
v 1 = - 10 i m/s,
v 2 = ( 20 * cos 40° ) i + ( 20 * sin 40° ) j =
= ( 20 * 0.766 ) i + ( 20 * 0.64278 ) j = ( 15.32 i + 12.855 j ) m/s
Δ v = ( 15.32 i + 12.855 j ) - ( - 10 i ) =
= 15.32 i + 12.855 j + 10 i = 25.32 i + 12.855 j
| Δv | = √ ( 25.32² + 12.855²) = √806.35 = 28.4 m/s
Imp = 0.2 kg * 28.4 m/s = 5.68 N-s
Answer: D ) 5.68 N-s. 
 
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When was the most gravitational potential energy stored between the model and Earth? Assume that the model's mass did not change
Lyrx [107]

Answer:

As the ball falls from C to E, potential energy is converted to kinetic energy. The velocity of the ball increases as it falls, which means that the ball attains its greatest velocity, and thus its greatest kinetic energy

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8 0
2 years ago
A 0.75 kg book is pushed across the table with an acceleration of 0.3 m/s2. What force is being applied to the
Brums [2.3K]

Answer:

\boxed {\boxed {\sf 0.225 \ Newtons}}

Explanation:

We are asked to find the force being applied to a book. According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F= ma

The mass of the book is 0.75 kilograms and the acceleration is 0.3 meters per square second. Substitute these values into the formula.

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  • a= 0.3 m/s²

F= 0.75 \ kg * 0.3 \ m/s^2

Multiply.

F =0.225 \ kg * m/s^2

1 kilogram meter per second squared is equal to 1 Newton. Therefore, our answer of 0.225 kilogram meters per second squared is equal to 0.225 Newtons.

F= 0.225  \ N

<u>0.225 Newtons of force</u> are applied to the book.

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1). both

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4). Earth

5). Earth

6). Venus (It would be pretty hard for US to mistake Earth for a star.)


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2 years ago
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