Answer:
T₂ = 2482.34 N
Explanation:
Equations of balance of forces
Look at the force diagram in the attached graph:
∑Fx=0
T₂cos α -T₁senα = 0 Equation (1)
∑Fy=0
T₁cos α +T₂sinα-W =0 Equation(2)
Data
m=539 kg
g= 9.8 m/s²
W= m*g= 539 kg* 9.8 m/s²= 5282.2 N
T₁ = 1.88T₂
Problem development
in the equation (1)
T₂cos α-(1.88T₂)senα = 0 We divided the equation by ( T₂cos α)
1 - (1.88)tanα = 0
tanα= 1/(1.88)
tanα= 0.5319
α = 28°
T₂=(1.88T₂)tanα
in the equation (2)
(1.88T₂)cos α+ T₂sinα - 5282.2 =0 We divided the equation by cos α
1.88T₂+ T₂tanα - 5282.2/cos α =0
1.88T₂+ T₂tan(28°) - 5282.2/(cos 28°) =0
1.88T₂+ (0.53)T₂- 5982.46 =0
(2.41)T₂ = 5982.46
T₂ = 5982.46/(2.41)
T₂ = 2482.34 N
Answer:
The average induced emf in the loop is 0.20 V
Explanation:
Given:
Radius of loop m
Magnetic field T
Change in time sec
According to the faraday's law,
Induced emf is given by
Where magnetic flux
( here
)
Where
We neglect minus sign because it's shows lenz law
V
Therefore, the average induced emf in the loop is 0.20 V