You put two ice cubes in a glass and fill the glass to the rim with water. As the ice melts, the water level remains the same.
Answer: Option D
<u>Explanation:</u>
As the ice is already in the water, and that has melted, there is no addition of volume into the glass. The water spills out if extra volume is added to the container. Hence, as there is no more volume added, there should be no change seen in the level of water.
The water level stays the same. This is because either it is a solid or liquid, the volume remains same. The volume of ice before melting is same as the volume of water, when melted into.
Answer:
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
Explanation:
Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution
As we know that the distance moved in one revolution is given as
also the time period of revolution for both will remain same as they move with the time period of carousel
Now we can say that the speed is given as
so Juan will have less tangential speed. so correct answer will be
Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.
Answer:
Δy= 5,075 10⁻⁶ m
Explanation:
The expression that describes the interference phenomenon is
d sin θ = (m + ½) λ
As the observation is on a distant screen
tan θ = y / x
tan θ= sin θ/cos θ
As in ethanes I will experience the separation of the vines is small and the distance to the big screen
tan θ = sin θ
Let's replace
d y / x = (m + ½) λ
The width of a bright stripe at the difference in distance
y₁ = (m + ½) λ x / d
m = 1
y₁ = 3/2 λ x / d
Let's use m = 1, we look for the following interference,
m = 2
y₂ = (2+ ½) λ x / d
The distance to the screen is constant x₁ = x₂ = x₀
The width of the bright stripe is
Δy = λ x / d (5/2 -3/2)
Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)
Δy= 5,075 10⁻⁶ m
Answer:
Organisms and their environment
Answer: 0.29 kN
Explanation:
We have the following data:
is the weight of the astronaut on Earth
is the free fall acceleration due gravity on Earth (directed downwards)
is the free fall acceleration due gravity on Zuton (directed downwards)
is the acceleration of the spaceship at litoff (directed upwards)
We have to find the <u>magnitude of the force</u> 
the space ship exerts on the astronaut.
Firstly, we have to know weight has a direct relation with the mass and the acceleration due gravity. In the case of Earth is:
(1)
Where 
is the mass of the atronaut.
Isolating :
(2)
(3)
(4)
Now that we know the mass of the astronaut, we can find its weight on Zuton:
(5)
(6)
(7)
Then, we can calculate the force the space ship exerts on the astronaut by the following equation:
(8)
Isolating :
(9)
(10)
(11)
Finally:
