What is the magnitude of the force between two parallel wires 43.0 m long and 5.9 cm apart, each carrying 35.0 A in the same dir
1 answer:
Answer:
The magnitude of the force between the two wires is
Explanation:
If there are two parallel rectilinear conductors through which two electric currents of the same direction I1 and I2 circulate,both conductors will generate a magnetic field on each other, giving rise to a force between them.
To calculate the value of this force, first, according to the law of Biot and Savart, the magnetic field produced by conductor 1 over 2 is obtained, which will be given by the equation:
B1= equation 1
B1: Magnetic field produced by conductor 1
= free space permeability
a= distance between wires
I1= current carrying wire 1
This magnetic field exerts on a segment L of the conductor 2 through which a current of intensity I2 circulates, a force equal to:
F1-2= I2*L*B1 Equation2
We replaced B1 of the equation 1 in the equation 2:
F1-2=
F1-2=
If we calculate the force exerted by conductor 2 on conductor 1 we would arrive at exactly the same value:
F2-1= F1-2
For this problem, the magnitude of the force between the two parallel cables that conduct current in the same direction is:
F1-2=F2-1=F
Wb/A.m
I1=I1=35A
L=43M
a =5.9 cm= m
Answer: The magnitude of the force between the two wires is
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Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45 )
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = - 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
26b) 66.7%
27) 500 N
Explanation:
26.a) In a two pulley system, the load is attached to one of the pulleys. The other pulley is attached to a fixed surface, as well as one end of the rope. The other end of the rope goes around moving pulley, then around the fixed pulley.
26.b) Mechanical advantage is the ratio between the forces:
MA = load force / effort force
Efficiency is the ratio between the work:
e = work done on load / work done by effort
Work is force times distance.
e = (F load × d load) / (F effort × d effort)
Rearranging:
e = (F load / F effort) × (d load / d effort)
e = MA × (d load / d effort)
In a two pulley system, the load moves half the distance of the effort. So the efficiency is:
e = (4/3) × (1/2)
e = 2/3
e = 66.7%
27) In a three pulley system, the load moves a third of the distance of the effort.
e = (F load / F effort) × (d load / d effort)
0.40 = (600 N / F) × (1/3)
F = 500 N
