Which sizes influence the adhesive and rolling friction?
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By using the coefficient of linear expansion, the increase in the length of the metal plate is by 0.015m and in the area is by 0.3074.
The rate of change in length of a metal per degree change in temperature is known as the coefficient of linear expansion.
Given:
Coefficient of linear expansion, α = 29 x /k
Length, L1 = 10m
T1 = 25℃
T2 = 78℃
ΔT = 78 – 25 = 53℃
To find:
Change in length (ΔL) and area (ΔA) of metal plate = ?
Formula:
ΔL = α L1 ΔT
ΔA = A1 2 α ΔT
Calculations:
ΔL = 29 x x 10 x 53
ΔL = 0.01537m
ΔA = 100 x 2 x 29 x x 53
ΔA = 0.3074
A2 = 100.3074
Result:
The increase in the length and area is by 0.015m and 0.3074 respectively.
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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
Δx=5.79*10⁻⁹ m
<u><em>5.79*10⁻⁹ m is the uncertainty in the position.</em></u>