Sorry if this answer is a bit late...
We know that the volume of the water is 0.5 cube meters.
The density of water is 1.000 g/m^3 (that's the real density of water... maybe typos?)
The density of ice is 0.900 g/m^3 (that's the approximate density of ice)
First, let's multiply the volume of water by the density of the water to get the mass, since we know that the mass does not change with the ice gets frozen.
0.5 * 1.000 = 0.500 g
Now, let's divide that by the density of ice to get the volume of the ice.
0.500 / 0.900 = 5/9 cube meters
≈ 0.556 cube meters.
The volume of the ice is 5/9 or 0.556 cube meters.
Have an awesome day! :)
Answer:
Final volume, V2 = 24.62 L
Explanation:
Given the following data;
Initial volume = 40 L
Initial pressure = 80 Pa
Final pressure = 130 Pa
To find the final volume V2, we would use Boyles' law.
Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.
Mathematically, Boyles law is given by;
Substituting into the equation, we have;




Final volume, V2 = 24.62 Liters
Answer:
i think its B
Explanation:
but check it before do it
Answer:
3.6 m
Explanation:


Also


Therefore, the minimum distance L you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J