The boiling point of water, or any liquid, varies according to the surrounding atmospheric pressure. A liquid boils, or begins turning to vapor, when its internal vapor pressure equals the atmospheric pressure.
<span>The measure of an acute angle is between 0 degrees and 90 degrees. It must be smaller than the perpendicular angle i.e., 90 degree. So, the answer of your question would be false.
In short, Your Answer would be "False"
Hope this helps!</span>
Answer:
the Voltages become negative instead of positive but the magnitude remains the same.
Explanation:
If we remove the positive charge by dragging it back to the box at the bottom, and drag a negative charge (blue) toward the middle of the screen, the Voltages become negative instead of positive but the magnitude remains the same.
Answer:
W = 2.3 10² 
Explanation:
The force of the weight is
W = m g
let's use the concept of density
ρ= m / v
the volume of a sphere is
V = 
π r³
V = π (1.0 10⁻³)³
V = 4.1887 10⁻⁹ m³
the density of water ρ = 1000 kg / m³
m = ρ V
m = 1000 4.1887 10⁻⁹
m = 4.1887 10⁻⁶ kg
therefore the out of gravity is
W = 4.1887 10⁻⁶ 9.8
W = 41.05 10⁻⁶ N
now let's look for the electric force
F_e = q E
F_e = 12 10⁻¹² 15000
F_e = 1.8 10⁻⁷ N
the relationship between these two quantities is
= 41.05 10⁻⁶ / 1.8 10⁻⁷
\frac{W}{F_e} = 2,281 10²
W = 2.3 10²
therefore the weight of the drop is much greater than the electric force
Answer:
318.3 nm
Explanation:
We approximate the circular film as a cylinder of height h and radius, r. Its volume V = πr²h. Since this volume equals the volume of the oil drop, the height of the circular film is thus h = V/πr²
V = 10⁻¹⁰ m³ and r = 10 m
Substituting into h, we have
h = 10⁻¹⁰ m³/π(10)²
= 0.3183 × 10⁻¹² m
= 3183 × 10⁻⁹ m
= 318.3 nm
