Answer:
All three pendulums will have the same angular frequencies.
Explanation:
For a simple pendulum, the time period using the approximation 
is expressed as:
The angular frequency 
is defined as
Since the angular frequency remains unaffected by the initial angle (valid strictly for small angle approximations), we deduce that the angular frequencies of the three pendulums are identical.
Answer
Ceres, Pluto, and Eris are categorized as DWARF PLANETS.
A) Remaining planetesimals formed within the frost line are referred to as ASTEROIDS.
B) METEORITES are fragments of asteroids that have landed on Earth.
C) COMETS are celestial objects that are often visible with their long tails.
D) COMETS are also planetesimals that were left over and originated in the region of the solar system dominated by the jovian planets.
E) Meteor showers are linked to debris from COMETS.
The response is affirmative .
An organism exhibiting the dominant phenotype can have two potential genetic combinations (for a gene with two alleles):
It might be homozygous dominant (WW) or
it can be heterozygous dominant (Ww), which is also known as a carrier
For instance, two black sheep can produce offspring with white wool if both are heterozygous dominant. In this scenario, both parents could pass on the recessive allele, resulting in their offspring inheriting the phenotype of white wool characterized by the genotype ww.
Answer:
F = 0.535 N
Explanation:
We will apply energy concepts, considering both the peak and the bottom of the path.
Top
Em₀ = U = mg y
Bottom
= K = ½ m v²
Emo =
mg y = ½ m v²
v = √ (2gy)
y = L - L cos θ
v = √ (2g L (1 - cos θ))
Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.
F = ma
a = v² / r
For the turning radius, the cable length is r = L.
F = m 2g (1 - cos θ)
Now, let's find the result.
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
Answer:
La magnitud del EMF es 0.00055 volts
Explanation:
El EMF inducido es proporcional al cambio en el flujo magnético según la ley de Faraday:
Como en nuestro caso hay solo un lazo de alambre, entonces N=1 y obtenemos:
Necesitamos expresar el flujo magnético dada la geometría del problema;
donde A es el área de la bobina que permanece constante con el tiempo, y B es el campo magnético que cambia con el tiempo. Por lo tanto, la ecuación para el EMF se convierte en:
