Answer:
1) λ = 0.413 m
, 2)v = 25,213 m / s
, 3) T = 0.216 N
, 4) m = 22.04 10-3 kg
Explanation:
1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related
λ = 2L / n n = 1, 2, 3 ...
In this case L = 0.62 m and n = 3
Let's calculate
λ = 2 0.62 / 3
λ = 0.413 m
2) the velocity related to wavelength and frequency
v = λ f
v = 0.413 61
v = 25,213 m / s
3) let's use the equation
v = √T /μ
T = v² μ
T = 25,213² 3.4 10⁻⁴
T = 0.216 N
4) the rope tension is proportional to the hanging weight
T-W = 0
T = W
W = m g
m = W / g
m = 0.216 / 9.8
m = 22.04 10-3 kg
5) n = 2
λ = 2 0.62 / 2
λ = 0.62 m
6) v = λ f
v = 0.62 61
v = 37.82 m / s
7) T = v² μ
T = 37.82² 3.4 10⁻⁴
T = 0.486 N
8) m = W / g
m = 0.486 / 9.8
m = 49.62 10⁻³ kg
9) n = 1
λ = 2 0.62
λ = 1.24 m
v = 1.24 61
v = 75.64 m / s
T = v² miu
T = 75.64² 3.4 10⁻⁴
T = 2.572 10⁻² N
m = 2.572 10⁻² / 9.8
m = 262.4 10⁻³ kg
I don't know
but a single substance is any sample of the known elements found in the periodic table of elements elements are made up of atoms of the same kind and cannot be decomposed by any chemical means into any other simpler elements.
Circular acceleration has 2 components, the centripetal and tangential acceleration.
The formula for calculating the centripetal acceleration of a particle
is:
= v^2/r
Where v is the velocity or instantaneous speed and r is the radius.
= (3^2)/6 = 1.5 m/s^2
Assuming that the particle stays on the same circular path, another
thing it can have is the tangential acceleration. While the centripetal
acceleration is constant, tangential acceleration depends on the location of
the particle in the circular path. So to answer A and B, yes we can have an
acceleration of 6 and 4 respectively.
Answer:
v₁ =√2gh, v₂ = v₁ /√2
Explanation:
Let's use the concepts of energy and work to analyze each case
hill without rubbing. Energy is conserved
starting point. Highest part
Em₀ = U = mg h
final point. Lower part
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v₁ =√2gh
rubbing hill
in this case the energy is not conserved because it is converted into work of the friction force, therefore the variation of the energy is the work of the friction
W = Em_{f} - Em₀
they indicate half of the initial mechanical energy is lost due to friction
W = ½ Em₀
we substitute
- ½ Em₀ = Em_{f} - Em₀
The negative sign is because the friction work always opposes the movement
Em_{f} = ½ Em₀
½ m v₂² = ½ m g h
v₂ = √½ √2gh
v₂ = v₁ /√2