Answer:
Q = M * (Cf + C * 100 + Cv)
Cf and Cf are heats of fusion and vaporization and C is the heat required to heat mass M of water 1 deg
Q = .001 kg ( 3.36 * E5 + 100 deg * 4200 + 2.26 * E6) J
Q = .001 kg ( 3.36 J / kg + 4.2 J / kg + 22.6 J /kg) * 10E5
Q = .001 kg * 30.2 * 10E5 J / kg = 3020 J
Answer:
3
Explanation:
It is the middle of the data set. I hope this helped
Answer:
Speed of an EM wave is different in air and in water.
Explanation:
The speed of an EM wave is different in air and water .
Since the frequency of any EM wave is its characteristic property, therefore wavelength and speed of EM wave Change on changing the medium of propogation.
Also the refractive index of any medium is defined as the ratio between light's speed in vacuum to the light's speed in any medium.
Also ,it can be define as speed of any EM wave in any medium = c/refractive index of medium.
Here, c is the speed of light, .
Answer:
5.23 C
Explanation:
The current in the wire is given by I = ε/R where ε = induced emf in the wire and R = resistance of wire.
Now, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = AΔB and A = area of loop and ΔB = change in magnetic field intensity = B₂ - B₁
B₁ = 0.670 T and B₂ = 0 T
ΔB = B₂ - B₁ = 0 - 0.670 T = - 0.670 T
A = πD²/4 where D = diameter of circular loop = 13.2 cm = 0.132 m
A = π(0.132 m)²/4 = 0.01368 m² = `1.368 × 10⁻² m²
ε = -ΔΦ/Δt = -AΔB/Δt = -1.368 × 10⁻² m² × (-0.670 T)/Δt= 0.9166 × 10⁻² Tm²/Δt
Now, the resistance R of the circular wire R = ρl/A' where ρ = resistivity of copper wire = 1.68 x 10⁻⁸ Ω.m, l = length of wire = πD and A' = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m
R = ρl/A' = 1.68 x 10⁻⁸ Ω.m × π × 0.132 m÷π(2.25 × 10⁻³ m)²/4 = 0.88704/5.0625 = 0.1752 × 10⁻² Ω = 1.752 × 10⁻³ Ω
So, I = ε/R = 0.9166 × 10⁻² Tm²/Δt1.752 × 10⁻³ Ω
IΔt = 0.9166 × 10⁻² Tm²/1.752 × 10⁻³ Ω = 0.5232 × 10 C
Since ΔQ = It = 5.232 C ≅ 5.23 C
So the charge is 5.23 C
Answer:
W = 1.8 J
Explanation:
The amount of work required to move the given charges can be found by using the following formula:
where,
W = Work done = ?
k = Colomb's constant = 9 x 10⁹ Nm²/C²
q₁ = magnitude of first charge = 6 μC = 6 x 10⁻⁶ C
q₂ = magnitude of second charge = 4 μC = 4 x 10⁻⁶ C
Δr = change in distance = 18 cm - 6 cm = 12 cm = 0.12 m
Therefore,
<u>W = 1.8 J</u>