Only two horizontal forces act on a 3.0 kg body that can move over a frictionless floor. one force is 9.0 n, acting due east, an
1 answer:
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Answer:
The magnitude of the electric field equal to 2.40 N/C at 1.1537m from the wire.
Explanation:
using Guass law,
(guessing that a cylinder of radius r and length L around wire such that wire is at centre )
E. A = qin / e0
E ( 2πr L ) = (1.56 x x L) / (8.85 x
)
E = (1.56 x ) / (2πr x 8.85 x
)
so 2.40 = (1.54 x ) / (2πr x 8.85 x
)
2.40 (6.284r) = 0.174 x 10²
15.0816r = 17.4
r = 1.1537m
Answer:
a) h = 53.8 m, b) h_minimum = 28 m, h_maximum = 63.3 m
Explanation:
a) For this exercise let's use Bernoulli's equation.
The subscript 1 is for the tank and the subscript for the building
P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂
In general, the water tanks are open to the atmosphere, so P1 = Patm, also the tanks are very large so the speed of the water surface is very small v₁=0 and as they give us the precious static, this it is when the keys are closed so the output velocity is zero, v₂= 0. The height of the floors in a building is y₂ = 12 m
we substitute in Bernoulli's equation
P_{atm} + 0 + ρ g h = P₂ + 0 + ρ g y₂
h =
h = + y₂
indicate that the value of ΔP = 410 10³ Pa
we calculate
h = 410 10³ / (1000 9.8) + 12
h = 53.8 m
b) ask for the height range for the minimum and maximum pressure
h = ΔP / rho g
minimum
h_minimum = 275 103/1000 9.8
h_minimum = 28 m
maximums
h_maximo = 620 103/1000 9.8
h_maximum = 63.3 m