Es el conjunto de longitudes de onda de todas las radiaciones electromagnéticas
Answer:
50% of it .
Explanation:
50% of it is illuminated by the Sun.
a) 1.57 m/s
The sock spins once every 2.0 seconds, so its period is
T = 2.0 s
Therefore, the angular velocity of the sock is
The linear speed of the sock is given by
where
is the angular velocity
r = 0.50 m is the radius of the circular path of the sock
Substituting, we find:
B) Faster
In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:
At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:
Therefore, the new linear speed would be:
And substituting,
So, we see that the linear speed has doubled.
Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.
Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.
Mathematically also potential energy is represented as
Potential energy= mgh
Where m is the mass of the pendulum.
g is the acceleration due to gravity
h is the height from the bottom z the ground.
At the bottom of the swing,the height is zero, hence the potential energy is also zero.
The kinetic energy is represented mathematically as
Kinetic energy= 1/2 mv^2
Where m is the mass of the pendulum
v is the velocity of the pendulum
At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.
Energy can neither be created e destroyed. It can only be transferred from one form to another. Implying this law and the above explainations we conclude that at the bottom of the pendulum,the potential energy=0 and the kinetic energy=294J as the entire potential energy is converted to kinetic energy at the bottom.
Answer
given,
current (I) = 16 mA
circumference of the circular loop (S)= 1.90 m
Magnetic field (B)= 0.790 T
S = 2 π r
1.9 = 2 π r
r = 0.3024 m
a) magnetic moment of loop
M= I A
M=
M=
M=
b) torque exerted in the loop
