Proton: Its electrical charge is +1, its mass is 1, it's found in the nucleus
Neutron: Its electrical charge is 0, its mass is 1, it's found in the nucleus
Electron: Its electrical charge is -1, its mass is negligible, it's found outside the nucleus
Explanation:
Atoms consist of three types of particles:
Learn more about atoms:
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Answer:
1.43 kg
Explanation:
We can solve the problem by using Newton's second law:
where
F is the net force on an object
m is its mass
a is its acceleration
For the squirrel in the problem, we have:
is the acceleration
F = 5 N is the net force acting on it
Solving for m, we find its mass:
Answer:
<em>0.228 nm</em>
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Explanation:
Atomic mass number of copper = 64
but an atomic mass unit = 1.66 x 10^-27 kg
therefore, the mass of the copper atom m = 64 x 1.66 x 10^-27 kg = 1.06 x 10^-25 kg
The number of atoms in this mass n = ρ/m
where ρ is the density of copper = 8.96 x 10^3 kg/m^3
==> n = (8.96 x 10^3)/(1.06 x 10^-25) = 8.45 x 10^28 atoms/m^3
We know that the volume occupied by this amount of atoms n =
where a is the lattice constant
equating, we have
8.45 x 10^28 =
a = 4.389 x 10^9
we also know that
d = 1/a
where d is the smallest distance between the two copper atom.
d = 1/(4.389 x 10^9) = 2.28 x 10^-10 m
==> <em>0.228 nm</em>
Answer:
1. λ = 2 L, 2. v = 2L f₁ , 3. v = √ T /μ², 4. μ = 2,287 10⁻³ kg / m , 5. Δv / v = 0.058 , 6. Δμ / μ = 0.12 , 7. Δ μ = 0.3 10⁻³ kg / m ,
8. μ = (2.3 ±0.3) 10⁻³ kg / m
Explanation:
The speed of a wave is
v = λ f 1
Where f is the frequency and λ the wavelength
The speed is given by the physical quantities of the system with the expression
v = √ T /μ² 2
1) The fundamental frequency of a string is when at the ends we have nodes and a maximum in the center, therefore this is
L = λ / 2
λ = 2 L
2) For this we substitute in equation 1
v = 2L f₁
3) let's clear from equation 2
The speed of a wave is
v = λ f₁
Where f is the frequency and Lam the wavelength
The speed is given by the physical quantities of the system with the expression
v = √ T /μ² 2
4) linear density is
μ = T / (2 L f₁)²
μ = 5.08 / (2 0.812 29.02)²
μ = 2,287 10⁻³ kg / m
We maintain three significant length figures, so the result is reduced to
μ = 2.29 10⁻³ kg / m
5) the speed of the wave is
v = 2 L f₁
The fractional uncertainty is
Δv / v = ΔL / L + Δf₁ / F₁
Δv / v = 0.02 / 0.812 + 1 / 29.02
Δv / v = 0.024 + 0.034
Δv / v = 0.058
6) the equation for linear density is
μ = T / (2 L f₁)²
Δμ / μ = 2 ΔL / L + 2Δf₁ / f₁
The tension is an exact value therefore its uncertainty is zero ΔT = 0
Δμ / μ = 2 0.02 / 0.812 + 2 1 / 29.02
Δμ / μ = 0.12
7) absolute uncertainty
Δ μ = μ
Δ μ = 0.12 2.29 10⁻³ kg / m
Δ μ = 0.3 10⁻³ kg / m
8)
μ = (2.3 ±0.3) 10⁻³ kg / m
Answer:
Δλ = 3*10⁻³ m.
Explanation:
where v is the speed, λ is the wavelength and f is the frequency.
⇒ Δλ = 3*10⁻³ m.