Answer:
A=0.199
Explanation:
We are given that
Mass of spring=m=450 g=
Where 1 kg=1000 g
Frequency of oscillation=
Total energy of the oscillation=0.51 J
We have to find the amplitude of oscillations.
Energy of oscillator=
Where 
=Angular frequency
A=Amplitude
Using the formula
Hence, the amplitude of oscillation=A=0.199
There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.
To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.
Where
A = Cross-sectional Area
Y = Young's modulus
= Coefficient of linear expansion for steel
= Temperature Raise
Our values are given as,
Replacing we have,
Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N
Answer :
The number of vacancies (per meter cube) = 5.778 × 10^22/m^3.
Explanation:
Given,
Atomic mass of silver = 107.87 g/mol
Density of silver = 10.35 g/cm^3
Converting to g/m^3,
= 10.35 g/cm^3 × 10^6cm^3/m^3
= 10.35 × 10^6 g/m^3
Avogadro's number = 6.022 × 10^23 atoms/mol
Fraction of lattice sites that are vacant in silver = 1 × 10^-6
Nag = (Na * Da)/Aag
Where,
Nag = Total number of lattice sites in Ag
Na = Avogadro's number
Da = Density of silver
Aag = Atomic weight of silver
= (6.022 × 10^23 × (10.35 × 10^6)/107.87
= 5.778 × 10^28 atoms/m^3
The number of vacancies (per meter cube) = 5.778 × 10^28 × 1 × 10^-6
= 5.778 × 10^22/m^3.
