All objects emit what radiation
1 answer:
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Answer:
0.0034 sec
Explanation:
L = initial length
T = initial time period = 2.51 s
Time period is given as
L = 1.56392 m
L' = new length
ΔT = Rise in temperature = 142 °C
α = coefficient of linear expansion = 19 x 10⁻⁶ °C
New length due to rise of temperature is given as
L' = L + LαΔT
L' = 1.56392 + (1.56392) (19 x 10⁻⁶) (142)
L' = 1.56814 m
T' = New time period
New time period is given as
T' = 2.5134 sec
Change in time period is given as
ΔT = T' - T
ΔT = 2.5134 - 2.51
ΔT = 0.0034 sec
Refer to the figure shown below.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
Answer:
Given that
Dry-bulb temperature(T) =24°C
Wet-bulb temperature(Tw) = 17°C
Pressure ,P = 1 atm
As we know that psychrometric chart are drawn at constant pressure.
From the diagram
ω= specific humidity
Lets take these two lines Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P
From chart at point P
a)
Specific humidity,ω = 0.00922 kg/kg
b)
The enthalpy ( h)
h=47.59 KJ/kg
c)
The relative humidity, RH
RH= 49.58 %
d)
Specific volume ,
v= 0.853 m³/kg
