Answer:
962291.57928 m²
Explanation:
= Pressure =
(full reflection)
I = Intensity = 
P = Power = 
c = Speed of light = 
M = Mass of Sun = 
m = Mass of ship = 1500 kg
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
Force of radiation is given by

This force will balance the gravitational force as stated in the question

The area of the must be 962291.57928 m²
Radio waves are the longest
Answer:
The mass of the rule is 56.41 g
Explanation:
Given;
mass of the object suspended at zero mark, m₁ = 200 g
pivot of the uniform meter rule = 22 cm
Total length of meter rule = 100 cm
0 22cm 100cm
-------------------------Δ------------------------------------
↓ ↓
200g m₂
Apply principle of moment
(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)
(200 g)(22 cm) = m₂(78 cm)
m₂ = (200 g)(22 cm) / (78 cm)
m₂ = 56.41 g
Therefore, the mass of the rule is 56.41 g
I would have to say B failed because I think I read something about it being only 2law not 3
Answer:
V_f = 287.04 mL
Explanation:
We are given the initial/original volume of the glycerine as 285 mL.
Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;
V_f = V_i (1 + βΔT)
Where;
V_f is the final volume
V_i is the original volume = 285 mL
β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)
Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C
Thus, plugging in relevant values;
V_f = 285(1 + (5.97 × 10^(-4) × 12))
V_f = 287.04 mL