Answer:
685.6 J
Explanation:
The latent heat of vaporization of ammonia is
L = 1371.2 kJ/kg
mass of ammonia, m = 0.0005 Kg
Heat = mass x latent heat of vaporization
H = 0.0005 x 1371.2
H = 0.686 kJ
H = 685.6 J
Thus, the amount of heat required to vaporize the ammonia is 685.6 J.
The magnitude of the second charge given that the first is –6×10¯⁶ C and is located 0.05 m away is +3.0×10¯⁶ C
<h3>Coulomb's law equation </h3>
F = Kq₁q₂ / r²
Where
- F is the force of attraction
- K is the electrical constant
- q₁ and q₂ are two point charges
- r is the distance apart
<h3>How to determine the second charge </h3>
- Charge 1 (q₁) = –6×10¯⁶ C
- Electric constant (K) = 9×10⁹ Nm²/C²
- Distance apart (r) = 0.05 m
- Force (F) = 65 N
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Divide both side by Kq₁
q₂ = Fr² / Kq₁
q₂ = (65 × 0.05²) / (9×10⁹ × 6×10¯⁶)
q₂ = +3.0×10¯⁶ C (since the force is attractive)
Learn more about Coulomb's law:
brainly.com/question/506926
Answer:
6.65×10⁻⁷ s
Explanation:
Speed of light in water = Total distance traveled in water/time
c = d/t ....................... Equation 1
Where c = speed of the light pulse in water, d = distance traveled in water, t = time.
also
c = v/n ..................... Equation 2
Where v = speed of light in air, n = refractive index of water
Substituting equation 2 into equation 1
v/n = d/t
making t the subject of the equation,
t = nd/v.................. Equation 3
Given: 150 m.
Constant: n = 1.33, v = 3.0×10⁸ m/s
Substitute into equation 3
t = 1.33(150)/3.0×10⁸
t = 6.65×10⁻⁷ seconds.
Hence the time = 6.65×10⁻⁷ s
