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Tcecarenko [31]
3 years ago
7

Controlled variables

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
7 0

Answer:

A control variable in scientific experimentation is an experimental element which is constant and unchanged throughout the course of the investigation.

Explanation:

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German physicist Werner Heisenberg related the uncertainty of an object's position (Δx) to the uncertainty in its velocity (Δv)Δ
laiz [17]

Answer:

5.79*10⁻⁹ m is the uncertainty in the position.

Explanation:

Heisenberg's uncertainty principle assumes that it is not possible to know exactly all the data regarding the behavior of particles. In other words, at the subatomic level, it is impossible to know at the same moment where a particle is, how it moves and what its speed is.

So, Heisenberg's Uncertainty Principle gives a relationship between the standard deviation of an object's position and its momentum.

Δp*Δx= h/(4π)

where

  • Δp the standard deviation of the object's momentum,
  • Δx the standard deviation of the object's position,
  • h=6.63*10⁻³⁴ J.s is the Planck's constant.

By definition, the momentum of the electron equals the product of its mass and velocity. So, being the mass constant, you can said:

Δp= m*Δv

Replacing in the expresion of the Heisenberg's Uncertainty Principle:

m*Δv*Δx= h/(4π)

Then you know:

  • m=9.11*10⁻³¹ kg
  • Δv=0.01*10⁶ m/s
  • h=6.63*10⁻³⁴ J.s= 6.63*10⁻³⁴ (N*m)*s=6.63*10⁻³⁴ [(kg*m*s⁻²)*m]*s= 6.63*10⁻³⁴ kg*m²*s⁻¹

Replacing:

*Δx=6.63*10⁻³⁴ kg*m²*s⁻¹/(4π)

Taking π=3.14 and solving:

Δx=\frac{6.63*10^{-34}  kg*m^{2} *s^{-1} }{4*3.14*9.11*10^{-31}  kg*0.01*10^{6}  m/s}

Δx=5.79*10⁻⁹ m

<u><em>5.79*10⁻⁹ m is the uncertainty in the position.</em></u>

3 0
4 years ago
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on it
Nadya [2.5K]

Answer:

  K_b = 78 J

Explanation:

For this exercise we can use the conservation of energy relations

starting point. Lowest of the trajectory

        Em₀ = K = ½ mv²

final point. When it is at tea = 50º

        Em_f = K + U

        Em_f = ½ m v_b² + m g h

where h is the height from the lowest point

        h = L - L cos 50

        Em_f = ½ m v_b² + mg L (1 - cos50)

energy be conserve

        Em₀ = Em_f

         ½ mv² = ½ m v_b² + mg L (1 - cos50)

         K_b = ½ m v_b² + mg L (1 - cos50)

let's calculate

          K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)

          K_b = 36 +42.0

          K_b = 78 J

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3 years ago
1. What is the gravitational force between two 4 kilogram masses separated by a distance of 5
Anon25 [30]
Yo, I assumed that this situation occurs on Earth otherwise the answer would be different.

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Is brainly working for you guys?
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Yeah whats it doing for you?

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