Answer:
0.0025H
Explanation:
I didn't come here to be part of this all I wanted is just information for my research
Answer:
E) True. The girl has a larger tangential acceleration than the boy.
Explanation:
In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.
Angular and linear quantities are related
v = w r
a = α r
the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m
as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different
v₁ = w 1,2 (boy)
v₂ = w 1.8 (girl)
whereby
v₂> v₁
reviewing the claims we have
a₁ = α 1,2
a₂ = α 1.8
a₂> a₁
A) False. Tangential velocity is different from zero
B) False angular acceleration is the same for both
C) False. It is the opposite, according to the previous analysis
D) False. Angular acceleration is equal
E) True. You agree with the analysis above,
Answer:
The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.
In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.
If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225
Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.
Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
