Answer:
The correct answer is option C. Sun.
Explanation:
The tail of a comet as well as the comma are visible thanks to the influence of sunlight on these objects.
If there is a case that these objects cross the inner Solar System,<u> then they will be visible from Earth</u>. The reason why this occurs is that as the comet approaches the internal solar system, the materials that are inside the comet vaporize and flow out of the nucleus thanks to solar radiation, and while they do carry dust with them.
This dust is the one that reflects sunlight directly, while the <u>gases shine because of ionization.
</u>
Usually we will need a telescope to see a comet, although there are some that can be seen with the naked eye.
Answer:
In this movement, both the magnitude and the direction of the force change
Explanation:
A body that moves in an elliptical path must be subjected to a force that is pointed towards one of the foci, therefore the force is central.
The magnitude of this force must be greater when the two bodies are closer and its directional changes, but it always points to the focus where the body that originates the force is.
In this movement, both the magnitude and the direction of the force change
Answer:
P(A∪B)=0.8
Explanation:
Given that,
A and B are mutually exclusive events such that,
P(A)=0.5 and P(B)=0.3
We need to find P(A or B). It means we need to find P(A∪B). If two events are mutually exclusive, P(A∩B)=0. So,
P(A∪B)=P(A)+P(B)-P(A∩B)
Putting P(A) and P(B), we get :
P(A∪B)=0.5+0.3-0
P(A∪B)=0.8
So, the value of P(A or B) is 0.8
Radio waves are reflected, absorbed, scattered, refracted, and diffracted by the atmospheric conditions that they encounter, such as clouds and precipitation. ... For example, only certain wavelengths pass through clouds unimpeded.
Answer:
E = 15 P₀
Explanation:
The power dissipated in a light bulb is
P = V I
V = I R
P = V² / R
the power is defined by
P = W / t
work equals energy
P = E / t
we substitute
V² / R = E / t
E = V² t / R
let's reduce the time to SI units
t = 1 min = 60 s
let's calculate the dissipated energy
In the exercise it does not indicate the nominal voltage of the bulb, but in general this voltage is V₀= 120 V
The applied voltage is half the nominal voltage
V = V₀ / 2
V = 120/2 = 60 V
E = (V₀ / 2)² t / R
E = ¼ t V₀² / R
E = ¼ 60 P₀
E = 15 P₀
Many times the nominal power (P₀) is written on the box of the bulb
