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san4es73 [151]
3 years ago
10

_______ says the universe began as a point and expanded.

Physics
1 answer:
Dominik [7]3 years ago
4 0
C Evolution says the universe began as a point and expanded
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Discuss how lawmakers could use scientific knowledge to help them make decisions that would benefit society.
Anika [276]
Scientific knowledge is very firm in its facts and pieces of evidence that create a more precise definition. If lawmakers are able to use the scientific method in making a decision, it will make a big change and will be beneficial to the society. The set of rules will be more arranged, specific and restrictive that might change justice, tax, and law.
7 0
3 years ago
A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When
katovenus [111]

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

\mathbf{f = \dfrac{v}{2L}}

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

\mathbf{254 - \dfrac{v}{2L} }

These combinations of frequency produce 4 beats per sound.

i.e.

\mathbf{254 - \dfrac{v}{2L}   =4}

\mathbf{ \dfrac{v}{2L} = 254-4 }

\mathbf{ \dfrac{v}{2L} = 250 ----(1)}

When it is altered, the beats first diminish and increase again by 4.

i.e.

\mathbf{ \dfrac{v'}{2L} = 254+4 }

\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }

If we equate both equations (1) and (2) together, we have:

\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;

  • when the temperature of the pipe  = unknown ???
  • the temperature of the open orang pipe = 15

∴

\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}

By squaring both sides, we have:

\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}

\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}

\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}

\implies \mathbf{273 +T =306.726912  }

T = 306.726912 - 273

T ≅ 33.73 ° C

∴

The change in temperature ΔT = 33.73° C - 15° C

The change in temperature ΔT = 18.73° C

Learn more about wave frequency here:

brainly.com/question/14316711?referrer=searchResults

4 0
3 years ago
Susie walks 3 blocks north to the local CVS store, then 4 blocks east to her grandmother’s house. She then walks 2 blocks west a
Slav-nsk [51]

Answer:

Suzie is 3 blocks north of where she started

Explanation:

Displacement is the minimum distance between the initial and final point of motion.

Here, Suzie first walks 3 blocks north. From there she walks 4 blocks east. Then 2 blocks to the east then 2 blocks north and then 2 blocks east. She covered 4 blocks east toward west. This is the same distance she covered traveling east. But she is 2 blocks north. From there she traveled a block south to the pizzeria and another block to her friends house. She covered the two block she had traveled north.

Hence, Suzie is 3 blocks north of where she started.

7 0
3 years ago
Far from any other masses, two masses, m1 and m2, are interacting gravitationally. The value for the mass of m1 suddenly doubles
NNADVOKAT [17]
<span>It also doubles The gravitational force between two masses is expressed as: F = G*m1*m2/r^2 where F = Force between the two masses m1 = Mass of object 1 m2 = Mass of object 2 r = distance between centers of object 1 and object 2 G = Gravitational constant The exact values of G, m1, m2, and r don't matter since all except for m1 is held constant. And when m1 suddenly doubles, the force attracting the two object to each other also doubles.</span>
5 0
4 years ago
A hockey puck oscillates on a frictionless, horizontal track while attached to a horizontal spring. The puck has mass 0.160 kg a
marshall27 [118]

Explanation:

The given data is as follows.

     mass (m) = 0.160 kg,            spring constant (k) = 8 n/m,

     Maximum speed (v_{m}) = 0.350 m/s

Formula for angular frequency is as follows.

          \omega = \sqrt{\frac{{k}{m}}

    \omega = \sqrt{\frac{{8}{0.160}}

    \omega = 7.07 rad/sec

(a) Formula to calculate the amplitude is as follows.

            \nu_{max} = A \omega

                  A = \frac{\nu}{\omega}

                      = \frac{0.35}{7.07}

                      = 0.05 m

Hence, value of amplitude is 0.05 m.

(b)   Displacement = 0.030 m

Formula for mechanical energy is as follows.

            M.E = \frac{1}{2}kA^{2}

Putting the values into the above formula as follows.

            M.E = \frac{1}{2}kA^{2}

                   = \frac{1}{2} \times 8 \times (0.05)^{2}

                   = 9.8 \times 10^{-3} Joule

For x = 0.03,

As,     P.E = \frac{1}{2} \times kx^{2}

                = \frac{1}{2} \times 8 \times (0.03)

                = 3.6 \times 10^{-3}

Hence, calculate the kinetic energy as follows.

            K.E = M.E - P.E

                  = (9.8 \times 10^{-3} - 3.6 \times 10^{-3}) J

                  = 6.2 \times 10^{-3} J

Thus, we can conclude that kinetic energy of the puck when the displacement of the glider is 0.0300 m is 6.2 \times 10^{-3} J.

7 0
3 years ago
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