A tank is 10 m long, 6 m wide, 3 m high, and contains kerosene with density 820 kg/m3 to a depth of 2.5 m. (Use 9.8 m/s2 for the
2 answers:
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Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
- L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
The magnitudes of the net magnetic fields at points A and B is 2.66 x T
Explanation:
Given information :
The current of each wires, I = 4.7 A
dH = 0.19 m
dV = 0.41 m
The magnetic of straight-current wire :
B= μI/2πr
where
B = magnetic field (T)
μ = 1.26 x
(N/
)
I = Current (A)
r = radius (m)
the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,
BH = μI/2πr
= (1.26 x )(4.7)/(2π)(0.19)
= 4.96 x T
BV = μI/2πr
= (1.26 x )(4.7)/(2π)(0.41)
= 2.3 x T
hence,
the net magnetic field = BH - BV
= 4.96 x - 2.3 x
= 2.66 x T