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just olya [345]
3 years ago
15

Which are functions of proteins?

Physics
2 answers:
NikAS [45]3 years ago
5 0
D. to be structural material
sweet-ann [11.9K]3 years ago
4 0
The answer is d to be a structural material
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What is conclusion about potential difference (voltmeter readings) in a parallel electric circuit ?
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Answer: the conclusion is that

Explanation:

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Atoms that are bonded together to form a new material with new<br> properties and characteristics. .
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Atoms form chemical bonds to make their outer electron shells more stable. ... An ionic bond, where one atom essentially donates an electron to another, forms when one atom becomes stable by losing its outer electrons and the other atoms become stable (usually by filling its valence shell) by gaining the electrons.

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How do protons, neutrons and electrons determine the properties of the atoms??? HELP! NEED SCIENCE HELP FAST!!!!
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4 0
3 years ago
A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre
Brut [27]

Answer:

The time taken is  t =  0.356 \ s

Explanation:

From the question we are told that

  The length of steel the wire is  l_1  = 31.0 \ m

   The  length of the  copper wire is  l_2  = 17.0 \ m

    The  diameter of the wire is  d =  1.00 \ m  =  1.0 *10^{-3} \ m

     The  tension is  T  =  122 \ N

     

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

              t  =  t_s  +  t_c

Where  t_s is the time taken to transverse the steel wire which is mathematically represented as

         t_s  = l_1 *  [ \sqrt{ \frac{\rho * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_s is the density of steel with a value  \rho_s  =  8920 \ kg/m^3

   So

      t_s  = 31 *  [ \sqrt{ \frac{8920 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_s  = 0.235 \ s

 And

        t_c is the time taken to transverse the copper wire which is mathematically represented as

      t_c  = l_2 *  [ \sqrt{ \frac{\rho_c * \pi *  d^2 }{ 4 *  T} } ]

here  \rho_c is the density of steel with a value  \rho_s  =  7860 \ kg/m^3

 So

      t_c  = 17 *  [ \sqrt{ \frac{7860 * 3.142*  (1*10^{-3})^2 }{ 4 *  122} } ]

      t_c  =0.121

So  

   t  = t_c  + t_s

    t =  0.121 + 0.235

    t =  0.356 \ s

4 0
3 years ago
A 2.0 kg bucket is attached to a horizontal ideal spring and rests on frictionless ice. You have a 1.0 kg mass
bogdanovich [222]

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

   

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = \sqrt{2y_{o}/g}

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = \sqrt{k/m}

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - \sqrt{k/3} \sqrt{2y_{o}/g}

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = \frac{\pi }{2} - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

8 0
3 years ago
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