m = mass of water in bathtub = 62.7 kg
= initial temperature of water in tub = 31 c
M = mass of water added = ?
= initial temperature of water added = 76 c
T = final equilibrium temperature of the system = 40.3 c
c = specific heat of water = 4186
Using conservation of heat
Heat gained by water in tub = Heat lost by water added
m c (T -
) = M c (
- T)
m (T -
) = M (
- T)
(62.7) (40.3 - 31) = M (76 - 40.3)
M = 16.3 kg
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Answer:
3.53×10⁶ N/c due west
Explanation:
From the question
E = F'/q........................ Equation 1
Where E = Electric Field, F = Net Force, q = Charge.
But,
F' = F₂-F₁...................... Equation 2
Substitute equation 2 into equation 1
E = (F₂-F₁)/q................ Equation 3
Given: F₁ = 3 N due east, F₂ = 15 N due west, q = 3.4×10⁻⁶ C
Substitute these values into equation 1
E = (15-3)/(3.4×10⁻⁶)
E = 12/(3.4×10⁻⁶)
E = 3.53×10⁶ N/c due west
Answer:
V_{average} =
, V_{average} = 2 V
Explanation:
he average or effective voltage of a wave is the value of the wave in a period
V_average = ∫ V dt
in this case the given volage is a square wave that can be described by the function
V (t) = 
to substitute in the equation let us separate the into two pairs
V_average = 
V_average = 
V_{average} = 
we evaluate V₀ = 4 V
V_{average} = 4 / 2)
V_{average} = 2 V