Answer:
Yes
Explanation:
Eclipses: Eclipses are also known as game of shadows where one object comes between the star(light source) and another object in a straight line such that the shadow of one object falls on other object. This can occur when the apparent size of the star and the object is almost same.
Talking about the Earth, the geometry is such that the Moon and the Sun are of same apparent size as seen from the Earth. Thus Lunar and Solar eclipse can be seen from the Earth. If we were to go on any other planet the same phenomenon can be seen provided the apparent size of moon and the Sun from that planet is same.
We have seen and recorded many such eclipses on Jupiter. These are from the perspective of Earth. When the moons of Jupiter comes exactly between the Sun and Jupiter the shadow of moon will fall on Jupiter. The places where the shadow falls, one will see a solar eclipse.
Answer:
3. Madison, WI
4. Quito, Equador
6. Main Sequence stars are becoming red giants
Explanation:
Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
Answer:
The answer is B.
Explanation:
Given that the <em>current </em>(Ampere) in a series circuit is same so we can ignore it. We can assume that the total voltage is 60V and all the 3 resistance are different, 20Ω, 40Ω and 60Ω. So first, we have to find the total resistance by adding :
Total resistance = 20Ω + 40Ω + 60Ω
= 120Ω
Next, we have to find out that 1Ω is equal to how many voltage by dividing :
120Ω = 60V
1Ω = 60V ÷ 120
1Ω = 0.5V
Lastly, we have to calculate the voltage at R1 so we have to multiply by 20 (R1) :
1Ω = 0.5V
20Ω = 0.5V × 20
20Ω = 10V
The kinetic energy of the electron is
where
is the mass of the electron and v its speed. Since we know the value of the kinetic energy,
, we can find the value of the speed v:
