Answer:
linear density of the string = 4.46 × 10⁻⁴ kg/m
Explanation:
given,
mass of the string = 31.2 g
length of string = 0.7 m
linear density of the string = 
linear density of the string = 
linear density of the string = 44.57 × 10⁻³ kg/m
linear density of the string = 4.46 × 10⁻⁴ kg/m
Answer:
4,524,660 N
Explanation:
Assuming the submarine's density is uniform, 1/9th of the submarine's mass is equal to the mass of the displaced water.
m/9 = (1026 kg/m³) (50 m³)
m = 461,700 kg
mg = 4,524,660 N
<span>The correct answer should be B) 63.55. That's because the most precise number is 63.546, but you would write 55 because 46 is rounded that way in the equation. The others are a bit higher, while E is a completely different element, Iodine. This isn't the most precise piece of data because in reality there would be a slight differentiation of +- 0,003u</span>
PH of 4 is Acidic and its property is to turn blue litmus red
