How could you create an investigation to determine the pitch of this wave?
1 answer:
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Answer:
the mass of steam at 100°C must be mixed is 150 g
Explanation:
given information:
ice's mass, = 490 g = 0.49 kg
steam temperature, T = 100°C
liquid water temperature, T = 89.0°C
specific heat of water, = 4186 J/kg.K = 4.186 kJ/kg.K
latent heat of fusion, = 333 kJ/kg
latent heat of vaporization, = 2256 kJ/kg
first, we calculate the heat of melted ice to water
Q₁ =
where
Q = heat
= mass of the ice
= latent heat of fusion
thus,
Q₁ =
= 0.49 x 333
= 163.17 kJ
then, the heat needed to increase the temperature of water to 89.0°C
Q₂ =
(89 - 0), the temperature of ice is 0°C
= specific heat of water
so,
Q₂ =
(89 - 0)
= 0.49 x 4.186 x 89
= 182.55 kJ
so, the heat absorbed by the ice is
Q = Q₁ + Q₂
= 163.17 + 182.55
= 345.72 kJ
the temperature of the steam is 100°C, so the mass of the steam is
Q = +
(100 - 89)
Q = (
+
(11))
= Q/ [
+
(11)]
= 345.72/ [2256 + (4.186 x 11)]
= 0.15 kg
= 150 g
Answer:
The answer is below
Explanation:
Given that:
x(t) = at – bt2+c
a) x(t) = at – bt2+c
Substituting a = 1.4 m/s, b = 0.06 m/s2 and c =50 m gives:
x(t) = 1.4t - 0.06t² + 50
At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
b) x(t) = 1.4t - 0.06t² + 50
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m
The average velocity (v) is given as:
c) x(t) = 1.4t - 0.06t² + 50
At t = 15, x(5) = 1.4(15) - 0.06(15)² + 50 = 57.5 m
At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m
The average velocity (v) is given as:
Answer:
A. 80.0microColoumbs
B.120.0 microcoloumbs
C.37.3v
Explanation:
See attached file
