Answer:48.52 kJ
Explanation:
Given
Resistance
temperature increases from 
Voltage=50 V
Heat given(H)
Where V=voltage applied
t=time
R=Resistance
Heat absorbed by water is
where
m=mass of water
c=specific heat of water
=change in temperature
Therefore 90-41.48=48.52 kJ is not absorbed by water and leaves the system into the surroundings.
Incomplete question as many data is missing.I have assumed value of charge and electric field.The complete question is here
A charge of 28 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 5.00×10⁴ V/m.
What work is done by the electric force when the charge moves a distance of 2.70 m at an angle of 45.0 degrees downward from the horizontal?
Answer:
Explanation:
Given data
Charge q=28 nC
Electric field E=5.00×10⁴ V/m.
Distance d=2.70 m
Angle α=45°
To find
Work done by electric force
Solution
Answer:
v = 2.029 m/s
Explanation:
Given
L = 84.0 cm ⇒ R = 0.5*L = 0.5*84 cm = 42 cm = 0.42 m
m₁ = 0.600 kg
m₂ = 0.200 kg
g = 9.8 m/s²
u₁ = u₂ = 0 m/s
v₁ = ?
v₂ = ?
Due to gravity, the bar oscillates and becomes vertical. The mass that occupies the lower position is the one with the highest torque. The one that reduces the potential energy (the system tends to the position of minimum energy). This is achieved if the mass that goes down is 0.6kg (that goes down 42cm) and the one that goes up is 0.2kg (goes up 42cm).
In this system mechanical energy is conserved, so we can match its value in the horizontal position with the one in the vertical.
then
Ei = Ki + Ui = 0.5*(m₁+m₂)*(0)² + (m₁+m₂)*9.8*(0) = 0 J
Ef = Kf + Uf
⇒ Kf = 0.5*(m₁+m₂)*v² = 0.5*(0.6+0.2)*v² = 0.4*v²
⇒ Uf = m₁*g*h₁ + m₂*g*h₂ = 0.6*9.8*(-0.42) + 0.2*9.8*0.42 = - 1.6464
⇒ Ef = Kf + Uf = 0.4*v² - 1.6464
Since
0 = 0.4*v² - 1.6464 ⇒ v = 2.029 m/s
v is the same value due to the wooden rod is pivoted about a horizontal axis through its center and the masses are on opposite ends.
v₁ = v₂ = v ⇒ ω₁*R₁ = ω₂*R₂ ⇒ ω₁*R = ω₂*R ⇒ ω₁ = ω₂ = ω
⇒ v = ω*R
