Answer:
False, because current flow has no effect in the weight of a conductor.
Explanation:
Current flow has no effect in the weight of a conductor. Thus, the weight on the scale will be the same when current is flowing and when current is not flowing in the copper tube.
Therefore, If a copper tube has current flowing to the right in the presence of a magnetic field going forward ("into the board"), we expect the weight on the scale to be the same whether current is flowing or not.
The AMOUNT of energy the ball has doesn't change. It's 294 joules in Darwin's hand, and it's still 294 joules when the ball hits the ground. It's all PE before he let's it go, and it steadily changes from PE to KE all the way down.
It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.
More and more PE turns into KE as the ball falls, all the way down.
When the ball hits the ground, it has no more PE left. All of its mechanical energy is then KE.
Comparative investigation is a research method that makes comparison across different countries of the world. Limitation : data set in different countries may be defined differently. Benefit : the method is flexible.
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
Answer:
F₄ = 29.819 N
Explanation:
Given
F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N
F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N
F₃ = (0 i + 0 j + 4 k) N
Then we have
F₁ + F₂ + F₃ + F₄ = 0
⇒ F₄ = - (F₁ + F₂ + F₃)
⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N
The magnitude of the force will be
F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N